Displaying 1-10 of 23 results found.
1, 1, 3, 11, 42, 165, 654, 2605, 10398, 41550, 166116, 664299, 2656866, 10626810, 42505932, 170021123, 680079282, 2720306730, 10881206124, 43524782946, 174099048684, 696396028620, 2785583782248, 11142334464693, 44569336530174, 178277343463830, 713109368541588
REFERENCES
H. Brocard, Query 4293, L'Intermédiaire des Mathématiciens, 23 (1916), 58-59. - N. J. A. Sloane, Mar 08 2022
FORMULA
a(0)=a(1)=1; a(2)=3; a(n) = 4*a(n-1) + ((-1)^(n-1)-3)/2 * a(floor((n-1)/2)). - Carl R. White, Oct 06 2017
MATHEMATICA
a[n_] := a[n] = If[n < 3, 1 + 2 Boole@ PrimeQ@ n, 4 a[n - 1] + ((-1)^(n - 1) - 3)/2*a[Floor[(n - 1)/2]]]; Array[a, 27, 0] (* Michael De Vlieger, Oct 06 2017 *)
Triangle read by rows, A101688 * (an infinite lower triangular matrix with A002083 as the main diagonal and the rest zeros).
+20
1
1, 0, 1, 0, 1, 1, 0, 0, 1, 2, 0, 0, 1, 2, 3, 0, 0, 0, 2, 3, 6, 0, 0, 0, 2, 3, 6, 11, 0, 0, 0, 0, 3, 6, 11, 22, 0, 0, 0, 0, 3, 6, 11, 22, 42, 0, 0, 0, 0, 0, 6, 11, 22, 42, 84, 0, 0, 0, 0, 0, 6, 11, 22, 42, 84, 165, 0, 0, 0, 0, 0, 0, 11, 22, 42, 84, 165, 330
COMMENTS
Row sums = A002083(n+1): (1, 1, 2, 3, 6, 11, 22, 42, 84, 165,...). Sum of n-th row terms = rightmost term of next row.
Eigensequence of triangle A101688 = A002083 starting with offset 1: (1, 1, 2, 3, 6, 11, 22, 42,...).
FORMULA
Equals M * Q, where M = triangle A101688, Q = an infinite lower triangular matrix with (1, 1, 1, 2, 3, 6, 11, 22, 42,...) as the main diagonal and the rest zeros.
EXAMPLE
First few rows of the triangle:
1;
0, 1;
0, 1, 1;
0, 0, 1, 2;
0, 0, 1, 2, 3;
0, 0, 0, 2, 3, 6;
0, 0, 0, 2, 3, 6, 11;
0, 0, 0, 0, 3, 6, 11, 22;
0, 0, 0, 0, 3, 6, 11, 22, 42;
0, 0, 0, 0, 0, 6, 11, 22, 42, 84;
0, 0, 0, 0, 0, 6, 11, 22, 42, 84, 165;
0, 0, 0, 0, 0, 0, 11, 22, 42, 84, 165, 330;
0, 0, 0, 0, 0, 0, 11, 22, 42, 84, 165, 330, 654;
0, 0, 0, 0, 0, 0, .0, 22, 42, 84, 165, 330, 654, 1308;
0, 0, 0, 0, 0, 0, .0, 22, 42, 84, 165, 330, 654, 1308, 2605;
...
n(n+Z(n)), where Z( ) is the Narayana-Zidek-Capell sequence ( A002083).
+20
0
0, 2, 6, 12, 24, 40, 72, 126, 240, 459, 940, 1936, 4104, 8671, 18508, 39300, 83616, 177055, 374652, 789811, 1662400, 3488877, 7309588, 15279406, 31886928, 66422275, 138157708, 286924599, 595102144, 1232672869
Jacobsthal sequence (or Jacobsthal numbers): a(n) = a(n-1) + 2*a(n-2), with a(0) = 0, a(1) = 1; also a(n) = nearest integer to 2^n/3.
(Formerly M2482 N0983)
+10
713
0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, 5726623061, 11453246123
COMMENTS
Don Knuth points out (personal communication) that Jacobsthal may never have seen the actual values of this sequence. However, Horadam uses the name "Jacobsthal sequence", such an important sequence needs a name, and there is a law that says the name for something should never be that of its discoverer. - N. J. A. Sloane, Dec 26 2020
Number of ways to tile a 3 X (n-1) rectangle with 1 X 1 and 2 X 2 square tiles.
Also, number of ways to tile a 2 X (n-1) rectangle with 1 X 2 dominoes and 2 X 2 squares. - Toby Gottfried, Nov 02 2008 Also a(n) counts each of the following four things: n-ary quasigroups of order 3 with automorphism group of order 3, n-ary quasigroups of order 3 with automorphism group of order 6, (n-1)-ary quasigroups of order 3 with automorphism group of order 2 and (n-2)-ary quasigroups of order 3. See the McKay-Wanless (2008) paper. - Ian Wanless, Apr 28 2008 Also the number of ways to tie a necktie using n + 2 turns. So three turns make an "oriental", four make a "four in hand" and for 5 turns there are 3 methods: "Kelvin", "Nicky" and "Pratt". The formula also arises from a special random walk on a triangular grid with side conditions (see Fink and Mao, 1999). - arne.ring(AT)epost.de, Mar 18 2001
Also the number of compositions of n + 1 ending with an odd part (a(2) = 3 because 3, 21, 111 are the only compositions of 3 ending with an odd part). Also the number of compositions of n + 2 ending with an even part (a(2) = 3 because 4, 22, 112 are the only compositions of 4 ending with an even part). - Emeric Deutsch, May 08 2001 Arises in study of sorting by merge insertions and in analysis of a method for computing GCDs - see Knuth reference.
Number of perfect matchings of a 2 X n grid upon replacing unit squares with tetrahedra (C_4 to K_4):
o----o----o----o...
| \/ | \/ | \/ |
| /\ | /\ | /\ |
Also the numerators of the reduced fractions in the alternating sum 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 + ... - Joshua Zucker, Feb 07 2002 Also, if A(n), B(n), C(n) are the angles of the n-orthic triangle of ABC then A(1) = Pi - 2*A, A(n) = s(n)*Pi + (-2)^n*A where s(n) = (-1)^(n-1) * a(n) [1-orthic triangle = the orthic triangle of ABC, n-orthic triangle = the orthic triangle of the (n-1)-orthic triangle]. - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), Jun 05 2002
Also the number of words of length n+1 in the two letters s and t that reduce to the identity 1 by using the relations sss = 1, tt = 1 and stst = 1. The generators s and t and the three stated relations generate the group S3. - John W. Layman, Jun 14 2002 Sums of pairs of consecutive terms give all powers of 2 in increasing order. - Amarnath Murthy, Aug 15 2002 Excess clockwise moves (over counterclockwise) needed to move a tower of size n to the clockwise peg is -(-1)^n*(2^n - (-1)^n)/3; a(n) is its unsigned version. - Wouter Meeussen, Sep 01 2002 Also the absolute value of the number represented in base -2 by the string of n 1's, the negabinary repunit. The Mersenne numbers ( A000225 and its subsequences) are the binary repunits. - Rick L. Shepherd, Sep 16 2002 Note that 3*a(n) + (-1)^n = 2^n is significant for Pascal's triangle A007318. It arises from a Jacobsthal decomposition of Pascal's triangle illustrated by 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = (7 + 35 + 1) + (1 + 35 + 7) + (21 + 21) = 43 + 43 + 42 = 3a(7) - 1; 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = (1 + 56 + 28) + (28 + 56 + 1) + (8 + 70 + 8) = 85 + 85 + 86 = 3a(8)+1. - Paul Barry, Feb 20 2003 Number of positive integers requiring exactly n signed bits in the nonadjacent form representation.
Equivalently, number of length-(n-1) words with letters {0, 1, 2} where no two consecutive letters are nonzero, see example and fxtbook link. - Joerg Arndt, Nov 10 2012 Counts walks between adjacent vertices of a triangle. - Paul Barry, Nov 17 2003 Every amphichiral rational knot written in Conway notation is a palindromic sequence of numbers, not beginning or ending with 1. For example, for 4 <= n <= 12, the amphichiral rational knots are: 2 2, 2 1 1 2, 4 4, 3 1 1 3, 2 2 2 2, 4 1 1 4, 3 1 1 1 1 3, 2 3 3 2, 2 1 2 2 1 2, 2 1 1 1 1 1 1 2, 6 6, 5 1 1 5, 4 2 2 4, 3 3 3 3, 2 4 4 2, 3 2 1 1 2 3, 3 1 2 2 1 3, 2 2 2 2 2 2, 2 2 1 1 1 1 2 2, 2 1 2 1 1 2 1 2, 2 1 1 1 1 1 1 1 1 2. For the number of amphichiral rational knots for n=2*k (k=1, 2, 3, ...), we obtain the sequence 0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, ... - Slavik Jablan, Dec 26 2003
a(n+2) counts the binary sequences of total length n made up of codewords from C = {0, 10, 11}. - Paul Barry, Jan 23 2004 Number of permutations with no fixed points avoiding 231 and 132.
The n-th entry (n > 1) of the sequence is equal to the 2,2-entry of the n-th power of the unnormalized 4 X 4 Haar matrix: [1 1 1 0 / 1 1 -1 0 / 1 1 0 1 / 1 1 0 -1]. - Simone Severini, Oct 27 2004 a(n) is the number of Motzkin (n+1)-sequences whose flatsteps all occur at level 1 and whose height is less than or equal to 2. For example, a(4) = 5 counts UDUFD, UFDUD, UFFFD, UFUDD, UUDFD. - David Callan, Dec 09 2004 If (m + n) is odd, then 3*(a(m) + a(n)) is always of the form a^2 + 2*b^2, where a and b both equal powers of 2; consequently every factor of (a(m) + a(n)) is always of the form a^2 + 2*b^2. - Matthew Vandermast, Jul 12 2003 Number of "0,0" in f_{n+1}, where f_0 = "1" and f_{n+1} = a sequence formed by changing all "1"s in f_n to "1,0" and all "0"s in f_n to "0,1". - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Sep 22 2006
All prime Jacobsthal numbers A049883[n] = {3, 5, 11, 43, 683, 2731, 43691, ...} have prime indices except for a(4) = 5. All prime Jacobsthal numbers with prime indices (all but a(4) = 5) are of the form (2^p + 1)/3 - the Wagstaff primes A000979[n]. Indices of prime Jacobsthal numbers are listed in A107036[n] = {3, 4, 5, 7, 11, 13, 17, 19, 23, 31, 43, 61, ...}. For n>1 A107036[n] = A000978[n] Numbers n such that (2^n + 1)/3 is prime. - Alexander Adamchuk, Oct 03 2006 Correspondence: a(n) = b(n)*2^(n-1), where b(n) is the sequence of the arithmetic means of previous two terms defined by b(n) = 1/2*(b(n-1) + b(n-2)) with initial values b(0) = 0, b(1) = 1; the g.f. for b(n) is B(x) := x/(1-(x^1+x^2)/2), so the g.f. A(x) for a(n) satisfies A(x) = B(2*x)/2. Because b(n) converges to the limit lim (1-x)*B(x) = 1/3*(b(0) + 2*b(1)) = 2/3 (for x --> 1), it follows that a(n)/2^(n-1) also converges to 2/3 (see also A103770). - Hieronymus Fischer, Feb 04 2006 Inverse: floor(log_2(a(n))) = n - 2 for n >= 2. Also: log_2(a(n) + a(n-1)) = n - 1 for n >= 1 (see also A130249). Characterization: x is a Jacobsthal number if and only if there is a power of 4 (= c) such that x is a root of p(x) = 9*x*(x-c) + (c-1)*(2*c+1) (see also the indicator sequence A105348). - Hieronymus Fischer, May 17 2007 This sequence counts the odd coefficients in the expansion of (1 + x + x^2)^(2^n - 1), n >= 0. - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Oct 18 2007, Jan 08 2008
2^(n+1) = 2* A005578(n) + 2*a(n) + 2* A000975(n-1). Let A005578(n), a(n), A000975(n-1) = triangle (a, b, c). Then ((S-c), (S-b), (S-a)) = ( A005578(n-1), a(n-1), A000975(n-2)). Example: (a, b, c) = (11, 11, 10) = ( A005578(5), a(5), A000975(4)). Then ((S-c), (S-b), (S-a)) = (6, 5, 5) = ( A005578(4), a(4), A000975(3)). - Gary W. Adamson, Dec 24 2007 Sequence is identical to the absolute values of its inverse binomial transform. A similar result holds for [0, A001045*2^n]. - Paul Curtz, Jan 17 2008 From a(2) on (i.e., 1, 3, 5, 11, 21, ...) also: least odd number such that the subsets of {a(2), ..., a(n)} sum to 2^(n-1) different values, cf. A138000 and A064934. It is interesting to note the pattern of numbers occurring (or not occurring) as such a sum ( A003158). - M. F. Hasler, Apr 09 2008 It appears that a(n) is also the number of integers between 2^n and 2^(n+1) that are divisible by 3 with no remainder. - John Fossaceca (john(AT)fossaceca.net), Jan 31 2009
Number of pairs of consecutive odious (or evil) numbers between 2^(n+1) and 2^(n+2), inclusive. - T. D. Noe, Feb 05 2009 A three-dimensional interpretation of a(n+1) is that it gives the number of ways of filling a 2 X 2 X n hole with 1 X 2 X 2 bricks. - Martin Griffiths, Mar 28 2009 Convolved with (1, 2, 2, 2, ...) = A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 23 2009 The product of a pair of successive terms is always a triangular number. - Giuseppe Ottonello, Jun 14 2009 Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := -2, A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = (-1)^(n-1)*det(A). - Milan Janjic, Jan 26 2010 Let R denote the irreducible representation of the symmetric group S_3 of dimension 2, and let s and t denote respectively the sign and trivial irreducible representations of dimension 1. The decomposition of R^n into irreducible representations consists of a(n) copies of R and a(n-1) copies of each of s and t. - Andrew Rupinski, Mar 12 2010 As a fraction: 1/88 = 0.0113636363... or 1/9898 = 0.00010103051121... - Mark Dols, May 18 2010 Starting with "1" = the INVERT transform of (1, 0, 2, 0, 4, 0, 8, ...); e.g., a(7) = 43 = (1, 1, 1, 3, 5, 11, 21) dot (8, 0, 4, 0, 2, 0, 1) = (8 + 4 + 10 + 21) = 43. - Gary W. Adamson, Oct 28 2010 Rule 28 elementary cellular automaton ( A266508) generates this sequence. - Paul Muljadi, Jan 27 2011 Let U be the unit-primitive matrix (see [Jeffery])
U = U_(6,2) =
(0 0 1)
(0 2 0)
(2 0 1).
Then a(n+1) = (Trace(U^n))/3, a(n+1) = ((U^n)_{3, 3})/3, a(n) = ((U^n)_{1, 3})/3 and a(n) = ((U^(n+1))_{1, 1})/2. (End)
The sequence emerges in using iterated deletion of strictly dominated strategies to establish the best-response solution to the Cournot duopoly problem as a strictly dominant strategy. The best response of firm 1 to firm 2's chosen quantity is given by q*_1 = 1/2*(a - c - q_2), where a is reservation price, c is marginal cost, and q_2 is firm two's chosen quantity. Given that q_2 is in [o, a - c], q*_1 must be in [o, 1/2*(a - c)]. Since costs are symmetric, we know q_2 is in [0, 1/2*(a - c)]. Then we know q*_1 is in [1/4*(a - c), 1/2*(a - c)]. Continuing in this way, the sequence of bounds we get (factoring out a - c) is {1/2, 1/4, 3/8, 5/16, ...}; the numerators are the Jacobsthal numbers. - Michael Chirico, Sep 10 2011 The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 3*a(n-1) equals the number of 3-colored compositions of n with all parts greater than or equal to 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011 This sequence is connected with the Collatz problem. We consider the array T(i, j) where the i-th row gives the parity trajectory of i, for example for i = 6, the infinite trajectory is 6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1... and T(6, j) = [0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, ..., 1, 0, 0, 1, ...]. Now, we consider the sum of the digits "1" of each column. We obtain the sequence a(n) = Sum_{k = 1..2^n} T(k, n) = Sum _{k = 1..2^n} digits "1" of the n-th column. Because a(n) + a(n+1) = 2^n, then a(n+1) = Number of digits "0" among the 2^n elements of the n-th column. - Michel Lagneau, Jan 11 2012 3!*a(n-1) is apparently the trace of the n-th power of the adjacency matrix of the complete 3-graph, a 3 X 3 matrix with diagonal elements all zero and off-diagonal all ones. The off-diagonal elements for the n-th power are all equal to a(n) while each diagonal element seems to be a(n) + 1 for an even power and a(n) - 1 for an odd. These are related to the lengths of closed paths on the graph (see Delfino and Viti's paper). - Tom Copeland, Nov 06 2012 2^n * a(-n) = (-1)^(n-1) * a(n), which extends the sequence to negative indices: ..., -5/16, 3/8, -1/4, 1/2, 0, 1, 1, 3, 5, ...
The "autosequence" property with respect to the binomial transform mentioned in my comment of Jan 17 2008 is still valid if the term a(-1) is added to the array of the sequence and its iterated higher-order differences in subsequent rows:
0 1/2 1/2 3/2 5/2 11/2 ...
1/2 0 1 1 3 5 ...
-1/2 1 0 2 2 6 ...
3/2 -1 2 0 4 4 ...
-5/2 3 -2 4 0 8 ...
11/2 -5 6 -4 8 0 ...
The main diagonal in this array contains 0's. (End)
Assign to a triangle T(n, 0) = 1 and T(n+1, 1) = n; T(r, c) = T(r-1, c-1) + T(r-1, c-2) + T(r-2, c-2). Then T(n+1, n) - T(n, n) = a(n). - J. M. Bergot, May 02 2013 a(n+1) counts clockwise walks on n points on a circle that take steps of length 1 and 2, return to the starting point after two full circuits, and do not duplicate any steps (USAMO 2013, problem 5). - Kiran S. Kedlaya, May 11 2013 Define an infinite square array m by m(n, 0) = m(0, n) = a(n) in top row and left column and m(i, j) = m(i, j-1) + m(i-1, j-1) otherwise, then m(n+1, n+1) = 3^(n-1). - J. M. Bergot, May 10 2013 a(n) is the number of compositions (ordered partitions) of n - 1 into one sort of 1's and two sorts of 2's. Example: the a(4) = 5 compositions of 3 are 1 + 1 + 1, 1 + 2, 1 + 2', 2 + 1 and 2' + 1. - Bob Selcoe, Jun 24 2013 Without 0, a(n)/2^n equals the probability that n will occur as a partial sum in a randomly-generated infinite sequence of 1's and 2's. The limiting ratio is 2/3. - Bob Selcoe, Jul 04 2013 Number of conjugacy classes of Z/2Z X Z/2Z in GL(2,2^(n+1)). - _Jared Warner_, Aug 18 2013
a(n) is the top left entry of the (n-1)-st power of the 3 X 3 matrix [1, 1, 1, 1, 0, 0, 1, 0, 0]. a(n) is the top left entry of the (n+1)-st power of any of the six 3 X 3 matrices [0, 1, 0; 1, 1, 1; 0, 1, 0], [0, 1, 1; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 1, 1; 1, 1, 0], [0, 1, 1; 1, 0, 1; 0, 1, 1], [0, 0, 1; 0, 0, 1; 1, 1, 1] or [0, 1, 0; 1, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014 This is the only integer sequence from the family of homogeneous linear recurrence of order 2 given by a(n) = k*a(n-1) + t*a(n-2) with positive integer coefficients k and t and initial values a(0) = 0 and a(1) = 1 whose ratio a(n+1)/a(n) converges to 2 as n approaches infinity. - Felix P. Muga II, Mar 14 2014 sqrt(a(n+1) * a(n-1)) -> a(n) + 3/4 if n is even, and -> a(n) - 3/4 if n is odd, for n >= 2. - Richard R. Forberg, Jun 24 2014 a(n+1) counts closed walks on the end vertices of P_3 containing one loop at the middle vertex. a(n-1) counts closed walks on the middle vertex of P_3 containing one loop at that vertex. - David Neil McGrath, Nov 07 2014 Let P be a point in the plane of a triangle ABC (with sides a, b, c) and barycentric coordinates P = [x:y:z]. The Complement of P with respect to ABC is defined to be Complement(P) = [b*y + c*z : c*z + a*x : a*x + b*y].
Then, for n >= 1, Complement(Complement(...(Complement(P))..)) = (n times) =
[2*a(n-1)*a*x + (2*a(n-1) - (-1)^n)*(b*y + c*z):
2*a(n-1)*b*y + (2*a(n-1) - (-1)^n)*(c*z + a*x):
2*a(n-1)*c*z + (2*a(n-1) - (-1)^n)*(a*x + b*y)]. (End)
a(n) (n >= 2) is the number of induced hypercubes of the Fibonacci cube Gamma(n-2). See p. 513 of the Klavzar reference. Example: a(5) = 11. Indeed, the Fibonacci cube Gamma(3) is <>- (cycle C(4) with a pendant edge) and the hypercubes are: 5 vertices, 5 edges, and 1 square. - Emeric Deutsch, Apr 07 2016 If the sequence of points {P_i(x_i, y_i)} on the cubic y = a*x^3 + b*x^2 + c*x + d has the property that the segment P_i(x_i, y_i) P_i+1(x_i+1, y_i+1) is always tangent to the cubic at P_i+1(x_i+1, y_i+1) then a(n) = -2^n*a/b*(x_(n+1)-(-1/2)^n*x_1). - Michael Brozinsky, Aug 01 2016 With the quantum integers defined by [n+1]_q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)), the Jacobsthal numbers are a(n+1) = (-1)^n*q^n [n+1]_q with q = i * sqrt(2) for i^2 = -1, whereas the signed Mersenne numbers A000225 are given by q = sqrt(2). Cf. A239473. - Tom Copeland, Sep 05 2016 Every positive integer has a unique expression as a sum of Jacobsthal numbers in which the index of the smallest summand is odd, with a(1) and a(2) both allowed. See the L. Carlitz, R. Scoville, and V. E. Hoggatt, Jr. reference. - Ira M. Gessel, Dec 31 2016. See A280049 for these expansions. - N. J. A. Sloane, Dec 31 2016 For n > 0, a(n) equals the number of ternary words of length n-1 in which 0 and 1 avoid runs of odd lengths. - Milan Janjic, Jan 08 2017 For n > 0, a(n) equals the number of orbits of the finite group PSL(2,2^n) acting on subsets of size 4 for the 2^n+1 points of the projective line. - Paul M. Bradley, Jan 31 2017 For n > 1, number of words of length n-2 over alphabet {1,2,3} such that no odd letter is followed by an odd letter. - Armend Shabani, Feb 17 2017 Also, the decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017 Also the number of independent vertex sets and vertex covers in the 2 X (n-2) king graph. - Eric W. Weisstein, Sep 21 2017 Let T(0) be a triangle and let T(1) be the medial triangle of T(0), T(2) the medial triangle of T(1) and, in general, T(n) the medial triangle of T(n-1). The barycentric coordinates of the first vertex of T(n) are [2*a(n-1)/a(n), 1, 1], for n > 0.
Let S(0) be a triangle and let S(1) be the antimedial triangle of S(0), S(2) the antimedial triangle of S(1) and, in general, S(n) the antimedial triangle of S(n-1). The barycentric coordinates of the first vertex of S(n) are [-a(n+1)/a(n), 1, 1], for n > 0. (End)
a(n) is also the number of derangements in S_{n+1} with empty peak set. - Isabella Huang, Apr 01 2018 For n > 0, gcd(a(n), a(n+1)) = 1. - Kengbo Lu, Jul 27 2020 Number of 2-compositions of n+1 with 1 not allowed as a part; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020 The number of Hamiltonian paths of the flower snark graph of even order 2n > 2 is 12*a(n-1). - Don Knuth, Dec 25 2020 When set S = {1, 2, ..., 2^n}, n>=0, then the largest subset T of S with the property that if x is in T, then 2*x is not in T, has a(n+1) elements. For example, for n = 4, #S = 16, a(5) = 11 with T = {1, 3, 4, 5, 7, 9, 11, 12, 13, 15, 16} (see Hassan Tarfaoui link, Concours Général 1991). - Bernard Schott, Feb 14 2022 a(n) is the number of words of length n over a binary alphabet whose position in the lexicographic order is one more than a multiple of three. a(3) = 3: aaa, abb, bba. - Alois P. Heinz, Apr 13 2022 Named by Horadam (1988) after the German mathematician Ernst Jacobsthal (1882-1965). - Amiram Eldar, Oct 02 2023 Define the sequence u(n) = (u(n-1) + u(n-2))/u(n-3) with u(0) = 0, u(1) = 1, u(2) = u(3) = -1. Then u(4*n) = -1 + (-1)^n/a(n+1), u(4*n+1) = 2 - (-1)^n/a(n+1), u(4*n+2) = u(4*n+3) = -1. For example, a(3) = 3 and u(8) = -2/3, u(9) = 5/3, u(10) = u(11) = -1. - Michael Somos, Oct 24 2023 Also, a(n) is the number of (3-color) states of a cycle (n+1)-pole C_{n+1} with n+1 terminals (or semiedges).
For instance, for n=3, the a(3)=3 states (3-coloring of the terminals) of C_4 are
a a a a a b
a a b b a b (End)
Also, with offset 1, the cogrowth sequence of the 6-element dihedral group D3. - Sean A. Irvine, Nov 04 2024
REFERENCES
Jathan Austin and Lisa Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:1 (2020), 340-350.
Thomas Fink and Yong Mao, The 85 ways to tie a tie, Fourth Estate, London, 1999; Die 85 Methoden eine Krawatte zu binden. Hoffmann und Kampe, Hamburg, 1999.
International Mathematical Olympiad 2001, Hong Kong Preliminary Selection Contest Problem #16.
Jablan S. and Sazdanovic R., LinKnot: Knot Theory by Computer, World Scientific Press, 2007. See p. 80.
Ernst Erich Jacobsthal, Fibonaccische Polynome und Kreisteilungsgleichungen, Sitzungsber. Berliner Math. Gesell. 17 (1919-1920), 43-57.
Tanya Khovanova, "Coins and Logic", Chapter 6, The Mathematics of Various Entertaining Subjects: Volume 3 (2019), Jennifer Beineke & Jason Rosenhouse, eds. Princeton University Press, Princeton and Oxford, p. 73. ISBN: 0691182582, 978-0691182582.
Donald E. Knuth, Art of Computer Programming, Vol. 3, Sect. 5.3.1, Eq. 13.
Thomas Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.
Steven Roman, Introduction to Coding and Information Theory, Springer Verlag, 1996, 41-42.
P. D. Seymour and D. J. A. Welsh, Combinatorial applications of an inequality form statistical mechanics, Math. Proc. Camb. Phil. Soc. 77 (1975), 485-495. [Although Daykin et al. (1979) claim that the present sequence is studied in this article, it does not seem to be explicitly mentioned. Note that definition of log-convex in (3.1) is wrong. - N. J. A. Sloane, Dec 26 2020] N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Robert M. Young, Excursions in Calculus, MAA, 1992, p. 239
LINKS
Paula Catarino, Helena Campos, and Paulo Vasco. On the Mersenne sequence. Annales Mathematicae et Informaticae, 46 (2016) pp. 37-53. S. Northshield, Stern's diatomic sequence 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, ..., Amer. Math. Monthly, 117 (2010), 581-598. USA Mathematical Olympiad 2013, Problem 2 (proposed by Sam Vandervelde). Eric Weisstein's World of Mathematics, King Graph. G. B. M. Zerr, Problem 64, American Mathematical Monthly, vol. 3, no. 12, 1896 (p. 311).
FORMULA
a(n) = 2^(n-1) - a(n-1). a(n) = 2*a(n-1) - (-1)^n = (2^n - (-1)^n)/3.
G.f.: x/(1 - x - 2*x^2) = x/((x+1)*(1-2*x)). Simon Plouffe in his 1992 dissertation E.g.f.: (exp(2*x) - exp(-x))/3.
a(2*n) = 2*a(2*n-1)-1 for n >= 1, a(2*n+1) = 2*a(2*n)+1 for n >= 0. - Lee Hae-hwang, Oct 11 2002; corrected by Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002 Also a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x, y) = x*F(n-1)(x, y) + y*F(n-2)(x, y), with y=2*x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*3^(k-1). - Paul Barry, Apr 02 2003 The ratios a(n)/2^(n-1) converge to 2/3 and every fraction after 1/2 is the arithmetic mean of the two preceding fractions. - Gary W. Adamson, Jul 05 2003 a(n) = U(n-1, i/(2*sqrt(2)))*(-i*sqrt(2))^(n-1) with i^2=-1. - Paul Barry, Nov 17 2003 a(n+1) = Sum_{k=0..ceiling(n/2)} 2^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004 a(n) = round(2^n/3) = (2^n + (-1)^(n-1))/3 so lim_{n->infinity} 2^n/a(n) = 3. - Gerald McGarvey, Jul 21 2004 a(n) = Sum_{k=0..n-1} (-1)^k*2^(n-k-1) = Sum_{k=0..n-1}, 2^k*(-1)^(n-k-1). - Paul Barry, Jul 30 2004 a(n+1) = Sum_{k=0..n} binomial(k, n-k)*2^(n-k). - Paul Barry, Oct 07 2004 a(n) = Sum_{k=0..n-1} W(n-k, k)*(-1)^(n-k)*binomial(2*k,k), W(n, k) as in A004070. - Paul Barry, Dec 17 2004 a(n) = Sum_{k=0..n} k*binomial(n-1, (n-k)/2)*(1+(-1)^(n+k))*floor((2*k+1)/3).
a(n+1) = Sum_{k=0..n} k*binomial(n-1, (n-k)/2)*(1+(-1)^(n+k))*( A042965(k)+0^k). (End) a(n+1) = ceiling(2^n/3) + floor(2^n/3) = (ceiling(2^n/3))^2 - (floor(2^n/3))^2.
a(n+1) = Sum_{k=0..n} Sum_{j=0..n} (-1)^(n-j)*binomial(j, k). - Paul Barry, Jan 26 2005 Let M = [1, 1, 0; 1, 0, 1; 0, 1, 1], then a(n) = (M^n)[2, 1], also matrix characteristic polynomial x^3 - 2*x^2 - x + 2 defines the three-step recursion a(0)=0, a(1)=1, a(2)=1, a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n > 2. - Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
a(n) = Sum_{k=0..floor(n/3)} binomial(n, f(n-1)+3*k);
a(n) = Sum_{k=0..floor(n/3)} binomial(n, f(n-2)+3*k), where f(n)= A080425(n). (End) a(2*n) = (1/3)*Product_{d|n} cyclotomic(d,4).
a(2*n+1) = (1/3)*Product_{d|2*n+1} cyclotomic(2*d,2). (End)
The a(n) are closely related to nested square roots; this is 2*sin(2^(-n)*Pi/2*a(n)) = sqrt(2-sqrt(2-sqrt(2-sqrt(...sqrt(2)))...) {with the '2' n times, n >= 0}.
Also 2*cos(2^(-n)*Pi*a(n)) = sqrt(2-sqrt(2-sqrt(2-sqrt(...sqrt(2)))...) {with the '2' n-1 times, n >= 1} as well as
2*sin(2^(-n)*3/2*Pi*a(n)) = sqrt(2+sqrt(2+sqrt(2+sqrt(...sqrt(2)))...) {with the '2' n times, n >= 0} and
2*cos(2^(-n)*3*Pi*a(n)) = -sqrt(2+sqrt(2+sqrt(2+sqrt(...sqrt(2)))...) {with the '2' n-1 times, n >= 1}.
a(n) = 2^(n+1)/Pi*arcsin(b(n+1)/2) where b(n) is defined recursively by b(0)=2, b(n)=sqrt(2-b(n-1)).
There is a similar formula regarding the arccos function, this is a(n) = 2^n/Pi*arccos(b(n)/2).
With respect to the sequence c(n) defined recursively by c(0)=-2, c(n)=sqrt(2+c(n-1)), the following formulas hold true: a(n) = 2^n/3*(1-(-1)^n*(1-2/Pi*arcsin(c(n+1)/2))); a(n) = 2^n/3*(1-(-1)^n*(1-1/Pi*arccos(-c(n)/2))).
(End)
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0,] = [ A005578(n), a(n), A000975(n-1)]. - Gary W. Adamson, Dec 24 2007 a(n) = Sum_{k=1..n} K(2, k)*a(n - k), where K(n,k) = k if 0 <= k <= n and K(n,k)=0 otherwise. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, the Fibonacci sequence can be generated in several ways using the K-coefficient.) - Thomas Wieder, Jan 13 2008 a(n) + a(n+2*k+1) = a(2*k+1)*2^n. - Paul Curtz, Feb 12 2008 a(n) = lower left term in the 2 X 2 matrix [0,2; 1,1]^n. - Gary W. Adamson, Mar 02 2008 a(n) = sqrt(8*a(n-1)*a(n-2) + 1). E.g., sqrt(3*5*8+1) = 11, sqrt(5*11*8+1) = 21. - Giuseppe Ottonello, Jun 14 2009
Let p[i] = Fibonacci(i-1) and let A be the Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, May 08 2010 Algebraically equivalent to replacing the 5's with 9's in the explicit (Binet) formula for the n-th term in the Fibonacci sequence: The formula for the n-th term in the Fibonacci sequence is F(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)). Replacing the 5's with 9's gives ((1+sqrt(9))^n - (1-sqrt(9))^n)/(2^n*sqrt(9)) = (2^n+(-1)^(n+1))/3 = (2^n-(-1)^(n))/3 = a(n). - Jeffrey R. Goodwin, May 27 2011 G.f.: x/(1-x-2*x^2) = G(0)/3; G(k) = 1 - ((-1)^k)/(2^k - 2*x*4^k/(2*x*2^k - ((-1)^k)/G(k+1))); (continued fraction 3 kind, 3-step).
E.g.f.: G(0)/3; G(k) = 1 - ((-1)^k)/(2^k - 2*x*4^k/(2*x*2^k - ((-1)^k)*(k+1)/G(k+1))); (continued fraction 3rd kind, 3-step). (End)
G.f.: Q(0)/3, where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 21 2013 G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(2*k+1 + 2*x)/( x*(2*k+2 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 29 2013 G.f.: Q(0) -1, where Q(k) = 1 + 2*x^2 + (k+2)*x - x*(k+1 + 2*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013 a(n) = (Sum_{k=1..n, k odd} C(n,k)*3^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014 a(-n) = -(-1)^n * a(n) / 2^n for all n in Z. - Michael Somos, Mar 18 2014 a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-3)^k = (2^n - (-1)^n)/3 = (-1)^(n-1)*Sum_{k=0..n-1} (-2)^k. Equals (-1)^(n-1)*Phi(n,-2), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014 a(2*n)/a(n) = A014551(n) for n >= 1; a(3*n)/a(n) = 3* A245489(n) for n >= 1. exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n.
exp( Sum_{n >= 1} a(3*n)/a(n)*x^n/n ) = Sum_{n >= 0} A084175(n+1)*x^n. exp( Sum_{n >= 1} a(4*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015266(n+3)*(-x)^n. exp( Sum_{n >= 1} a(5*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015287(n+4)*x^n. exp( Sum_{n >= 1} a(6*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015305(n+5)*(-x)^n. exp( Sum_{n >= 1} a(7*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015323(n+6)*x^n. exp( Sum_{n >= 1} a(8*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015338(n+7)*(-x)^n. exp( Sum_{n >= 1} a(9*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015356(n+8)*x^n. exp( Sum_{n >= 1} a(10*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015371(n+9)*(-x)^n. (End) Dirichlet g.f.: (PolyLog(s,2) + (1 - 2^(1-s))*zeta(s))/3. - Ilya Gutkovskiy, Jun 27 2016 a(m)*a(n) + a(m-1)*a(n-1) - 2*a(m-2)*a(n-2) = 2^(m+n-3).
a(m+n-1) = a(m)*a(n) + 2*a(m-1)*a(n-1); a(m+n) = a(m+1)*a(n+1) - 4*a(m-1)*a(n-1).
a(2*n-1) = a(n)^2 + 2*a(n-1)^2; a(2*n) = a(n+1)^2 - 4*a(n-1)^2. (End)
a(n+4) = a(n) + 5*2^n, a(0) = 0, a(1..4) = [1,1,3,5]. That is to say, for n > 0, the ones digits of Jacobsthal numbers follow the pattern 1,1,3,5,1,1,3,5,1,1,3,5,.... - Yuchun Ji, Apr 25 2019 The sequence starting with "1" is the second INVERT transform of (1, -1, 3, -5, 11, -21, 43, ...). - Gary W. Adamson, Jul 08 2019 a(n)^2 - a(n+1)*a(n-1) = (-2)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-2)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-2)^n*a(m-n).
a(n) = Sum_{i=0..n-1; j=0..n-1; i+2*j=n-1} 2^j*((i+j)!/(i!*j!)). (End)
For n > 0, 1/(2*a(n+1)) = Sum_{m>=n} a(m)/(a(m+1)*a(m+2)). - Kai Wang, Mar 03 2020 For 4 > h >= 0, k >= 0, a(4*k+h) mod 5 = a(h) mod 5. - Kai Wang, May 07 2020 a(n) = 1 + Sum_{k=0..n-1} a(k) if n odd; a(n) = Sum_{k=0..n-1} a(k) if n even.
a(n) = F(n) + Sum_{k=0..n-2} a(k)*F(n-k-1), where F denotes the Fibonacci numbers.
a(n) = b(n) + Sum_{k=0..n-1} a(k)*b(n-k), where b(n) is defined through b(0) = 0, b(1) = 1, b(n) = 2*b(n-2).
a(n) = 1 + 2*Sum_{k=0..n-2} a(k).
a(m+n) = a(m)*a(n+1) + 2*a(m-1)*a(n).
a(2*n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)*2^(i+j). (End)
G.f.: x/(1 - x - 2*x^2) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 2*x)/(1 + k*x) (a telescoping series). - Peter Bala, May 08 2024 a(n) = Sum_{r>=0} F(n-2r, r), where F(n, 0) is the n-th Fibonacci number and F(n,r) = Sum_{j=1..n} F(n+1-j, r-1) F(j, r-1). - Gregory L. Simay, Aug 31 2024
EXAMPLE
a(2) = 3 because the tiling of the 3 X 2 rectangle has either only 1 X 1 tiles, or one 2 X 2 tile in one of two positions (together with two 1 X 1 tiles).
The a(6)=21 length-5 ternary words with no two consecutive letters nonzero are (dots for 0's)
[ 1] [ . . . . ]
[ 2] [ . . . 1 ]
[ 3] [ . . . 2 ]
[ 4] [ . . 1 . ]
[ 5] [ . . 2 . ]
[ 6] [ . 1 . . ]
[ 7] [ . 1 . 1 ]
[ 8] [ . 1 . 2 ]
[ 9] [ . 2 . . ]
[10] [ . 2 . 1 ]
[11] [ . 2 . 2 ]
[12] [ 1 . . . ]
[13] [ 1 . . 1 ]
[14] [ 1 . . 2 ]
[15] [ 1 . 1 . ]
[16] [ 1 . 2 . ]
[17] [ 2 . . . ]
[18] [ 2 . . 1 ]
[19] [ 2 . . 2 ]
[20] [ 2 . 1 . ]
[21] [ 2 . 2 . ]
(End)
G.f. = x + x^2 + 3*x^3 + 5*x^4 + 11*x^5 + 21*x^6 + 43*x^7 + 85*x^8 + 171*x^9 + ...
MATHEMATICA
Jacob0[n_] := (2^n - (-1)^n)/3; Table[Jacob0[n], {n, 0, 33}] (* Robert G. Wilson v, Dec 05 2005 *) Array[(2^# - (-1)^#)/3 &, 33, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
LinearRecurrence[{1, 2}, {0, 1}, 40] (* Harvey P. Dale, Nov 30 2011 *) CoefficientList[Series[x/(1 - x - 2 x^2), {x, 0, 34}], x] (* Robert G. Wilson v, Jul 21 2015 *) Table[Abs[QBinomial[n, 1, -2]], {n, 0, 35}] (* John Keith, Jan 29 2022 *)
PROG
(PARI) a(n) = (2^n - (-1)^n) / 3
(PARI) M=[1, 1, 0; 1, 0, 1; 0, 1, 1]; for(i=0, 34, print1((M^i)[2, 1], ", ")) \\ Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
(Sage) [lucas_number1(n, 1, -2) for n in range(34)] # Zerinvary Lajos, Apr 22 2009 # Alternatively:
a = BinaryRecurrenceSequence(1, 2)
(Haskell)
a001045 = (`div` 3) . (+ 1) . a000079
a001045_list = 0 : 1 :
zipWith (+) (map (2 *) a001045_list) (tail a001045_list)
(Maxima)
a[0]:0$
a[n]:=2^(n-1)-a[n-1]$
(PARI) a=0; for(n=0, 34, print1(a, ", "); a=2*(a-n%2)+1) \\ K. Spage, Aug 22 2014 a, b = 0, 1
while True:
yield a
a, b = b, b+2*a
(Magma) [n le 2 select n-1 else Self(n-1)+2*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jun 27 2016 (Python) [(2**n-(-1)**n)//3 for n in range(40)] # Gennady Eremin, Mar 03 2022
CROSSREFS
Partial sums of this sequence give A000975, where there are additional comments from B. E. Williams and Bill Blewett on the tie problem. a(n) = A073370(n-1, 0), n >= 1 (first column of triangle). Cf. A000978, A000979, A019322, A066845, A105348, A130249, A130250, A130253, A005578, A002083, A113405, A138000, A064934, A003158, A175286 (Pisano periods), A147613, A156319, A002605, A000225, A052129, A014551 (companion "autosequence"), A015266, A015287, A015305, A015323, A015338, A015356, A015371, A084175, A245489, A283641.
EXTENSIONS
Thanks to Don Knuth, who pointed out several missing references, including Brocard (1880), which although it was mentioned in the 1973 Handbook of Integer Sequences, was omitted from the 1995 "Encyclopedia". - N. J. A. Sloane, Dec 26 2020
Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1).
(Formerly M0141 N0056)
+10
378
0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 6, 5, 9, 4, 11, 7, 10, 3, 11, 8, 13, 5, 12, 7, 9, 2, 9, 7, 12, 5, 13, 8, 11, 3, 10, 7, 11, 4, 9, 5, 6, 1, 7, 6, 11, 5, 14, 9, 13, 4, 15, 11, 18, 7, 17, 10, 13, 3, 14, 11, 19, 8, 21, 13, 18, 5, 17, 12, 19
COMMENTS
Also called fusc(n) [Dijkstra].
a(n)/a(n+1) runs through all the reduced nonnegative rationals exactly once [Stern; Calkin and Wilf].
If the terms are written as an array:
column 0 1 2 3 4 5 6 7 8 9 ...
row 0: 0
row 1: 1
row 2: 1,2
row 3: 1,3,2,3
row 4: 1,4,3,5,2,5,3,4
row 5: 1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5
row 6: 1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,...
...
then (ignoring row 0) the sum of the k-th row is 3^(k-1), each column is an arithmetic progression and the steps are nothing but the original sequence. - Takashi Tokita (butaneko(AT)fa2.so-net.ne.jp), Mar 08 2003
The above observation can be made more precise. Let A(n,k), n >= 0, 0 <= k <= 2^(n-1)-1 for k > 0, denote the entry in row n and column k of the left-justified array above.
The equations for columns 0,1,2,3,4,... are successively (ignoring row 0):
1 (n >= 1),
n (n >= 2),
n-1 (n >= 3),
2n-3 (n >= 3),
n-2 (n >= 4),
3n-7 (n >= 4),
...
and in general column k > 0 is given by
A(n,k) = a(k)*n - A156140(k) for n >= ceiling(log_2(k+1))+1, and 0 otherwise. (End)
a(n) is the number of odd Stirling numbers S_2(n+1, 2r+1) [Carlitz].
Moshe Newman proved that the fraction a(n+1)/a(n+2) can be generated from the previous fraction a(n)/a(n+1) = x by 1/(2*floor(x) + 1 - x). The successor function f(x) = 1/(floor(x) + 1 - frac(x)) can also be used.
a(n+1) = number of alternating bit sets in n [Finch].
If f(x) = 1/(1 + floor(x) - frac(x)) then f(a(n-1)/a(n)) = a(n)/a(n+1) for n >= 1. If T(x) = -1/x and f(x) = y, then f(T(y)) = T(x) for x > 0. - Michael Somos, Sep 03 2006 a(n+1) is the number of ways of writing n as a sum of powers of 2, each power being used at most twice (the number of hyperbinary representations of n) [Carlitz; Lind].
a(n+1) is the number of partitions of the n-th integer expressible as the sum of distinct even-subscripted Fibonacci numbers (= A054204(n)), into sums of distinct Fibonacci numbers [Bicknell-Johnson, theorem 2.1]. a(n+1) is the number of odd binomial(n-k, k), 0 <= 2*k <= n. [Carlitz], corrected by Alessandro De Luca, Jun 11 2014 a(2^k) = 1. a(3*2^k) = a(2^(k+1) + 2^k) = 2. Sequences of terms between a(2^k) = 1 and a(2^(k+1)) = 1 are palindromes of length 2^k-1 with a(2^k + 2^(k-1)) = 2 in the middle. a(2^(k-1) + 1) = a(2^k - 1) = k+1 for k > 1. - Alexander Adamchuk, Oct 10 2006 It appears that the terms of this sequence are the number of odd entries in the diagonals of Pascal's triangle at 45 degrees slope. - Javier Torres (adaycalledzero(AT)hotmail.com), Aug 06 2009
Let M be an infinite lower triangular matrix with (1, 1, 1, 0, 0, 0, ...) in every column shifted down twice:
1;
1, 0;
1, 1, 0;
0, 1, 0, 0;
0, 1, 1, 0, 0;
0, 0, 1, 0, 0, 0;
0, 0, 1, 1, 0, 0, 0;
...
Member of the infinite family of sequences of the form a(n) = a(2*n); a(2*n+1) = r*a(n) + a(n+1), r = 1 for A002487 = row 1 in the array of A178239. - Gary W. Adamson, May 23 2010 Equals row 1 in an infinite array shown in A178568, sequences of the form a(2*n) = r*a(n), a(2*n+1) = a(n) + a(n+1); r = 1. - Gary W. Adamson, May 29 2010 Row sums of A125184, the Stern polynomials. Equivalently, B(n,1), the n-th Stern polynomial evaluated at x = 1. - T. D. Noe, Feb 28 2011 The Kn1y and Kn2y triangle sums, see A180662 for their definitions, of A047999 lead to the sequence given above, e.g., Kn11(n) = A002487(n+1) - A000004(n), Kn12(n) = A002487(n+3) - A000012(n), Kn13(n) = A002487(n+5) - A000034(n+1) and Kn14(n) = A002487(n+7) - A157810(n+1). For the general case of the knight triangle sums see the Stern-Sierpiński triangle A191372. This triangle not only leads to Stern's diatomic series but also to snippets of this sequence and, quite surprisingly, their reverse. - Johannes W. Meijer, Jun 05 2011 Maximum of terms between a(2^k) = 1 and a(2^(k+1)) = 1 is the Fibonacci number F(k+2). - Leonid Bedratyuk, Jul 04 2012 Probably the number of different entries per antidiagonal of A223541. That would mean there are exactly a(n+1) numbers that can be expressed as a nim-product 2^x*2^y with x + y = n. - Tilman Piesk, Mar 27 2013 Let f(m,n) be the frequency of the integer n in the interval [a(2^(m-1)), a(2^m-1)]. Let phi(n) be Euler's totient function ( A000010). Conjecture: for all integers m,n n<=m f(m,n) = phi(n). - Yosu Yurramendi, Sep 08 2014 Define a sequence chf(n) of Christoffel words over an alphabet {-,+}: chf(1) = '-'; chf(2*n+0) = negate(chf(n)); chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))). Then the length of the chf(n) word is fusc(n) = a(n); the number of '-'-signs in the chf(n) word is c-fusc(n) = A287729(n); the number of '+'-signs in the chf(n) word is s-fusc(n) = A287730(n). See examples below. - I. V. Serov, Jun 01 2017 The sequence can be extended so that a(n) = a(-n), a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1) for all n in Z. - Michael Somos, Jun 25 2019 Named after the German mathematician Moritz Abraham Stern (1807-1894), and sometimes also after the French clockmaker and amateur mathematician Achille Brocot (1817-1878). - Amiram Eldar, Jun 06 2021
REFERENCES
M. Aigner and G. M. Ziegler, Proofs from THE BOOK, 3rd ed., Berlin, Heidelberg, New York: Springer-Verlag, 2004, p. 97.
Elwyn R. Berlekamp, John H. Conway and Richard K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 114.
Krishna Dasaratha, Laure Flapan, Chansoo Lee, Cornelia Mihaila, Nicholas Neumann-Chun, Sarah Peluse and Matthew Stroegeny, A family of multi-dimensional continued fraction Stern sequences, Abtracts Amer. Math. Soc., Vol. 33 (#1, 2012), #1077-05-2543.
Edsger W. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232 (sequence is called fusc).
F. G. M. Eisenstein, Eine neue Gattung zahlentheoretischer Funktionen, welche von zwei Elementen abhaengen und durch gewisse lineare Funktional-Gleichungen definirt werden, Verhandlungen der Koenigl. Preuss. Akademie der Wiss. Berlin (1850), pp. 36-42, Feb 18, 1850. Werke, II, pp. 705-711.
Graham Everest, Alf van der Poorten, Igor Shparlinski and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.16.3; pp. 148-149.
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 117.
Thomas Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Bruce Bates and Toufik Mansour, The q-Calkin-Wilf tree, Journal of Combinatorial Theory Series A, Vol. 118, No. 3 (2011), pp. 1143-1151. Aviezri S. Fraenkel, Ratwyt, Dec 28 2011. Brian Hayes, On the Teeth of Wheels, American Scientist, Vol. 88, No. 4 (July-August 2000), pp. 296-300 (5 pages). Sam Northshield, Stern's diatomic sequence 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, ..., Amer. Math. Monthly, Vol. 117, No. 7 (2010), pp. 581-598. Bruce Reznick, Some binary partition functions, in "Analytic number theory" (Conf. in honor P. T. Bateman, Allerton Park, IL, 1989), Progr. Math., 85, Birkhäuser Boston, Boston, MA, 1990, pp. 451-477.
FORMULA
a(n+1) = (2*k+1)*a(n) - a(n-1) where k = floor(a(n-1)/a(n)). - David S. Newman, Mar 04 2001 Let e(n) = A007814(n) = exponent of highest power of 2 dividing n. Then a(n+1) = (2k+1)*a(n)-a(n-1), n > 0, where k = e(n). Moreover, floor(a(n-1)/a(n)) = e(n), in agreement with D. Newman's formula. - Dragutin Svrtan (dsvrtan(AT)math.hr) and Igor Urbiha (urbiha(AT)math.hr), Jan 10 2002 Calkin and Wilf showed 0.9588 <= limsup a(n)/n^(log(phi)/log(2)) <= 1.1709 where phi is the golden mean. Does this supremum limit = 1? - Benoit Cloitre, Jan 18 2004. Coons and Tyler show the limit is A246765 = 0.9588... - Kevin Ryde, Jan 09 2021 a(n) = Sum_{k=0..floor((n-1)/2)} (binomial(n-k-1, k) mod 2). - Paul Barry, Sep 13 2004 a(n) = Sum_{k=0..n-1} (binomial(k, n-k-1) mod 2). - Paul Barry, Mar 26 2005 G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = v^3 + 2*u*v*w - u^2*w. - Michael Somos, May 02 2005 G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u1^3*u6 - 3*u1^2*u2*u6 + 3*u2^3*u6 - u2^3*u3. - Michael Somos, May 02 2005 G.f.: x * Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))) [Carlitz].
a(n) = a(n-2) + a(n-1) - 2*(a(n-2) mod a(n-1)). - Mike Stay, Nov 06 2006 a(n) = Sum_{k=1..n} K(k, n-k)*a(n - k), where K(n,k) = 1 if 0 <= k AND k <= n AND n-k <= 2 and K(n,k) = 0 else. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, if we drop the condition k <= n in the above definition, then we arrive at A002083 = Narayana-Zidek-Capell numbers.) - Thomas Wieder, Jan 13 2008 a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1; a(2^n - k) + a(k) = a(2^(n+1) + k). Both formulas hold for 0 <= k <= 2^n - 1. G.f.: G(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ... Define f(z) = (1 + z + z^2), then G(z) = lim f(z)*f(z^2)*f(z^4)* ... *f(z^(2^n))*... = (1 + z + z^2)*G(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 11 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1 (0 <= k <= 2^n - 1). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
a(2^n + k) = a(2^n - k) + a(k) (0 <= k <= 2^n). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
Let g(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ..., f(z) = 1 + z + z^2. Then g(z) = lim_{n->infinity} f(z)*f(z^2)*f(z^4)*...*f(z^(2^n)), g(z) = f(z)*g(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
For 0 <= k <= 2^n - 1, write k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1) where b(0), b(1), etc. are 0 or 1. Define a 2 X 2 matrix X(m) with entries x(1,1) = x(2,2) = 1, x(1,2) = 1 - b(m), x(2,1) = b(m). Let P(n)= X(0)*X(1)* ... *X(n-1). The entries of the matrix P are members of the sequence: p(1,1) = a(k+1), p(1,2) = a(2^n - (k+1)), p(2,1) = a(k), p(2,2) = a(2^n - k). - Arie Werksma (werksma(AT)tiscali.nl), Apr 20 2008
Let f(x) = A030101(x); if 2^n + 1 <= x <= 2^(n + 1) and y = 2^(n + 1) - f(x - 1) then a(x) = a(y). - Arie Werksma (Werksma(AT)Tiscali.nl), Jul 11 2008 Equals infinite convolution product of [1,1,1,0,0,0,0,0,0] aerated A000079 - 1 times, i.e., [1,1,1,0,0,0,0,0,0] * [1,0,1,0,1,0,0,0,0] * [1,0,0,0,1,0,0,0,1]. - Mats Granvik and Gary W. Adamson, Oct 02 2009; corrected by Mats Granvik, Oct 10 2009 a(k+1)*a(2^m+k) - a(k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014 a(2^(m+1)+(k+1))*a(2^m+k) - a(2^(m+1)+k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014 a(5*2^k) = 3. a(7*2^k) = 3. a(9*2^k) = 4. a(11*2^k) = 5. a(13*2^k) = 5. a(15*2^k) = 4. In general: a((2j-1)*2^k) = A007306(j), j > 0, k >= 0 (see Adamchuk's comment). - Yosu Yurramendi, Mar 05 2016 a(2^m+2^m'+k') = a(2^m'+k')*(m-m'+1) - a(k'), m >= 0, m' <= m-1, 0 <= k' < 2^m'. - Yosu Yurramendi, Jul 13 2016 Let n be a natural number and [b_m b_(m-1) ... b_1 b_0] its binary expansion with b_m=1.
Let L = Sum_{i=0..m} b_i be the number of binary digits equal to 1 (L >= 1).
Let {m_j: j=1..L} be the set of subindices such that b_m_j = 1, j=1..L, and 0 <= m_1 <= m_2 <= ... <= m_L = m.
If L = 1 then c_1 = 1, otherwise let {c_j: j=1..(L-1)} be the set of coefficients such that c_(j) = m_(j+1) - m_j + 1, 1 <= j <= L-1.
Let f be a function defined on {1..L+1} such that f(1) = 0, f(2) = 1, f(j) = c_(j-2)*f(j-1) - f(j-2), 3 <= j <= L+1.
Then a(n) = f(L+1) (see example). (End)
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + 2^m + k) = 2*a(k), m >= 0, 0 <= k < 2^m.
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + k) = a(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^m + k) = a(k)*(m - floor(log_2(k)) - 1) + a(2^(floor(log_2(k))+1) + k), m >= 0, 0 < k < 2^m, a(2^m) = 1, a(0) = 0. (End)
a(2^m) = 1, m >= 0.
a(2^r*(2*k+1)) = a(2^r*(2*k)) + a(2^r*(2*k+2)), r < - m - floor(log_2(k)) - 1, m > 0, 1 <= k < 2^m. (End)
Trow(n) = [card({k XOR j-k): k=0..j}) for j = 2^(n-1)-1..2^n-2] when regarded as an irregular table (n >= 1). - Peter Luschny, Sep 29 2024
EXAMPLE
Stern's diatomic array begins:
1,1,
1,2,1,
1,3,2,3,1,
1,4,3,5,2,5,3,4,1,
1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5,1,
1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,7,11,4,9,...
...
a(91) = 19, because 91_10 = 1011011_2; b_6=b_4=b_3=b_1=b_0=1, b_5=b_2=0; L=5; m_1=0, m_2=1, m_3=3, m_4=4, m_5=6; c_1=2, c_2=3, c_3=2, c_4=3; f(1)=1, f(2)=2, f(3)=5, f(4)=8, f(5)=19. - Yosu Yurramendi, Jul 13 2016 a(n) is the length of the Christoffel word chf(n):
1 '-' 1 1
2 '+' 2 1
3 '+-' 2 2
4 '-' 3 1
5 '--+' 3 3
6 '-+' 3 2
... (End)
G.f. = x + x^2 + 2*x^3 + x^4 + 3*x^5 + 2*x^6 + 3*x^7 + x^8 + ... - Michael Somos, Jun 25 2019
MAPLE
A002487 := proc(n) option remember; if n <= 1 then n elif n mod 2 = 0 then procname(n/2); else procname((n-1)/2)+procname((n+1)/2); fi; end: seq( A002487(n), n=0..91);
A002487 := proc(m) local a, b, n; a := 1; b := 0; n := m; while n>0 do if type(n, odd) then b := a+b else a := a+b end if; n := floor(n/2); end do; b; end proc: seq( A002487(n), n=0..91); # Program adapted from E. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232. - Igor Urbiha (urbiha(AT)math.hr), Oct 28 2002. Since A007306(n) = a(2*n+1), this program can be adapted for A007306 by replacing b := 0 by b := 1.
A002487 := proc(n::integer) local k; option remember; if n = 0 then 0 elif n=1 then 1 else add(K(k, n-1-k)*procname(n - k), k = 1 .. n) end if end proc:
K := proc(n::integer, k::integer) local KC; if 0 <= k and k <= n and n-k <= 2 then KC:=1; else KC:= 0; end if; end proc: seq( A002487(n), n=0..91); # Thomas Wieder, Jan 13 2008 # next Maple program:
a:= proc(n) option remember; `if`(n<2, n,
(q-> a(q)+(n-2*q)*a(n-q))(iquo(n, 2)))
end:
fusc := proc(n) local a, b, c; a := 1; b := 0;
for c in convert(n, base, 2) do
if c = 0 then a := a + b else b := a + b fi od;
b end:
Stern := proc(n, u) local k, j, b;
b := j -> nops({seq(Bits:-Xor(k, j-k), k = 0..j)}):
ifelse(n=0, 1-u, seq(b(j), j = 2^(n-1)-1..2^n-1-u)) end:
seq(print([n], Stern(n, 1)), n = 0..5); # As shown in the comments.
seq(print([n], Stern(n, 0)), n = 0..5); # As shown in the examples. # Peter Luschny, Sep 29 2024
MATHEMATICA
a[0] = 0; a[1] = 1; a[n_] := If[ EvenQ[n], a[n/2], a[(n-1)/2] + a[(n+1)/2]]; Table[ a[n], {n, 0, 100}] (* end of program *)
Onemore[l_] := Transpose[{l, l + RotateLeft[l]}] // Flatten;
NestList[Onemore, {1}, 5] // Flatten (*gives [a(1), ...]*) (* Takashi Tokita, Mar 09 2003 *)
ToBi[l_] := Table[2^(n - 1), {n, Length[l]}].Reverse[l]; Map[Length,
Split[Sort[Map[ToBi, Table[IntegerDigits[n - 1, 3], {n, 500}]]]]] (*give [a(1), ...]*) (* Takashi Tokita, Mar 10 2003 *)
A002487[m_] := Module[{a = 1, b = 0, n = m}, While[n > 0, If[OddQ[n], b = a+b, a = a+b]; n = Floor[n/2]]; b]; Table[ A002487[n], {n, 0, 100}] (* Jean-François Alcover, Sep 06 2013, translated from 2nd Maple program *)
a[0] = 0; a[1] = 1;
Flatten[Table[{a[2*n] = a[n], a[2*n + 1] = a[n] + a[n + 1]}, {n, 0, 50}]] (* Horst H. Manninger, Jun 09 2021 *) nmax = 100; CoefficientList[Series[x*Product[(1 + x^(2^k) + x^(2^(k+1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 08 2022 *)
PROG
(PARI) {a(n) = n=abs(n); if( n<2, n>0, a(n\2) + if( n%2, a(n\2 + 1)))};
(PARI) fusc(n)=local(a=1, b=0); while(n>0, if(bitand(n, 1), b+=a, a+=b); n>>=1); b \\ Charles R Greathouse IV, Oct 05 2008 (PARI) A002487(n, a=1, b=0)=for(i=0, logint(n, 2), if(bittest(n, i), b+=a, a+=b)); b \\ M. F. Hasler, Feb 12 2017, updated Feb 14 2019 (Haskell)
a002487 n = a002487_list !! n
a002487_list = 0 : 1 : stern [1] where
stern fuscs = fuscs' ++ stern fuscs' where
fuscs' = interleave fuscs $ zipWith (+) fuscs $ (tail fuscs) ++ [1]
interleave [] ys = ys
interleave (x:xs) ys = x : interleave ys xs
(R)
N <- 50 # arbitrary
a <- 1
for (n in 1:N)
{
a[2*n ] = a[n]
a[2*n + 1] = a[n] + a[n+1]
a
}
a
(Scheme)
;; An implementation of memoization-macro definec can be found for example in: http://oeis.org/wiki/Memoization
(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
def a(n): return n if n<2 else a(n//2) if n%2==0 else a((n - 1)//2) + a((n + 1)//2) # Indranil Ghosh, Jun 08 2017; corrected by Reza K Ghazi, Dec 27 2021 (Python)
def a(n):
a, b = 1, 0
while n > 0:
if n & 1:
b += a
else:
a += b
n >>= 1
return b
(Sage)
M = [1, 0]
for b in n.bits():
M[b] = M[0] + M[1]
return M[1]
print([ A002487(n) for n in (0..91)]) (Julia)
using Nemo
function A002487List(len)
a, A = QQ(0), [0, 1]
for n in 1:len
a = next_calkin_wilf(a)
push!(A, denominator(a))
end
A end
(R) # Given n, compute a(n) by taking into account the binary representation of n
a <- function(n){
b <- as.numeric(intToBits(n))
l <- sum(b)
m <- which(b == 1)-1
d <- 1
if(l > 1) for(j in 1:(l-1)) d[j] <- m[j+1]-m[j]+1
f <- c(0, 1)
if(l > 1) for(j in 3:(l+1)) f[j] <- d[j-2]*f[j-1]-f[j-2]
return(f[l+1])
(R) # computes the sequence as a vector A, rather than function a() as above.
A <- c(1, 1)
maxlevel <- 5 # by choice
for(m in 1:maxlevel) {
A[2^(m+1)] <- 1
for(k in 1:(2^m-1)) {
r <- m - floor(log2(k)) - 1
A[2^r*(2*k+1)] <- A[2^r*(2*k)] + A[2^r*(2*k+2)]
}}
(Magma) [&+[(Binomial(k, n-k-1) mod 2): k in [0..n]]: n in [0..100]]; // Vincenzo Librandi, Jun 18 2019 (Python)
def A002487(n): return sum(int(not (n-k-1) & ~k) for k in range(n)) # Chai Wah Wu, Jun 19 2022
CROSSREFS
Cf. A000123, A000360, A001045, A002083, A011655, A020950, A026741, A037227, A046815, A070871, A070872, A071883, A073459, A084091, A101624, A126606, A174980, A174981, A178239, A178568, A212288, A213369, A260443, A277020, A277325, A287729, A287730, A293160. If the 1's are replaced by pairs of 1's we obtain A049456. Cf. A049455 for the 0,1 version of Stern's diatomic array. Cf. A086592 and references therein to other sequences related to Kepler's tree of fractions.
Triangle of Narayana numbers T(n,k) = C(n-1,k-1)*C(n,k-1)/k with 1 <= k <= n, read by rows. Also called the Catalan triangle.
+10
369
1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1, 1, 15, 50, 50, 15, 1, 1, 21, 105, 175, 105, 21, 1, 1, 28, 196, 490, 490, 196, 28, 1, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 1, 55, 825, 4950, 13860, 19404, 13860, 4950, 825
COMMENTS
Number of antichains (or order ideals) in the poset 2*(k-1)*(n-k) or plane partitions with rows <= k-1, columns <= n-k and entries <= 2. - Mitch Harris, Jul 15 2000 T(n,k) is the number of Dyck n-paths with exactly k peaks. a(n,k) = number of pairs (P,Q) of lattice paths from (0,0) to (k,n+1-k), each consisting of unit steps East or North, such that P lies strictly above Q except at the endpoints. - David Callan, Mar 23 2004 Number of permutations of [n] which avoid-132 and have k-1 descents. - Mike Zabrocki, Aug 26 2004 T(n,k) is the number of paths through n panes of glass, entering and leaving from one side, of length 2n with k reflections (where traversing one pane of glass is the unit length). - Mitch Harris, Jul 06 2006 Antidiagonal sums given by A004148 (without first term). T(n,k) is the number of full binary trees with n internal nodes and k-1 jumps. In the preorder traversal of a full binary tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump. - Emeric Deutsch, Jan 18 2007 The n-th row can be generated by the following operation using an ascending row of (n-1) triangular terms, (A) and a descending row, (B); e.g., row 6:
A: 1....3....6....10....15
B: 15...10....6.....3.....1
C: 1...15...50....50....15....1 = row 6.
Leftmost column of A,B -> first two terms of C; then followed by the operation B*C/A of current column = next term of row C, (e.g., 10*15/3 = 50). Continuing with the operation, we get row 6: (1, 15, 50, 50, 15, 1). (End)
The previous comment can be upgraded to: The ConvOffsStoT transform of the triangular series; and by rows, row 6 is the ConvOffs transform of (1, 3, 6, 10, 15). Refer to triangle A117401 as another example of the ConvOffsStoT transform, and OEIS under Maple Transforms. - Gary W. Adamson, Jul 09 2012 T(n,k) is also the number of order-decreasing and order-preserving mappings (of an n-element set) of height k (height of a mapping is the cardinal of its image set). - Abdullahi Umar, Aug 21 2008 Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p.60]. See A033282 for the corresponding array of f-vectors for associahedra of type A_n. See A008459 and A145903 for the h-vectors for associahedra of type B and type D respectively. The Hilbert transform of this triangle (see A145905 for the definition of this transform) is A145904. - Peter Bala, Oct 27 2008 T(n,k) is also the number of noncrossing set partitions of [n] into k blocks. Given a partition P of the set {1,2,...,n}, a crossing in P are four integers [a, b, c, d] with 1 <= a < b < c < d <= n for which a, c are together in a block, and b, d are together in a different block. A noncrossing partition is a partition with no crossings. - Peter Luschny, Apr 29 2011 Noncrossing set partitions are also called genus 0 partitions. In terms of genus-dependent Stirling numbers of the second kind S2(n,k,g) that count partitions of genus g of an n-set into k nonempty subsets, one has T(n,k) = S2(n,k,0). - Robert Coquereaux, Feb 15 2024 Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = 1/sqrt(y)*BesselI(1,2*sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!*(n+1)! as defined in Wang and Wang.
Generating function E(y)*E(x*y) = 1 + (1 + x)*y/(1!*2!) + (1 + 3*x + x^2)*y^2/(2!*3!) + (1 + 6*x + 6*x^2 + x^3)*y^3/(3!*4!) + .... Cf. A105278 with a generating function exp(y)*E(x*y). The n-th power of this array has a generating function E(y)^n*E(x*y). In particular, the matrix inverse A103364 has a generating function E(x*y)/E(y). (End) T(n,k) is the number of nonintersecting n arches above the x axis, starting and ending on vertices 1 to 2n, with k being the number of arches starting on an odd vertice and ending on a higher even vertice. Example: T(3,2)=3 [16,25,34] [14,23,56] [12,36,45]. - Roger Ford, Jun 14 2014 Fomin and Reading on p. 31 state that the rows of the Narayana matrix are the h-vectors of the associahedra as well as its dual. - Tom Copeland, Jun 27 2017 The row polynomials P(n, x) = Sum_{k=1..n} T(n, k)*x^(k-1), together with P(0, x) = 1, multiplied by (n+1) are the numerator polynomials of the o.g.f.s of the diagonal sequences of the triangle A103371: G(n, x) = (n+1)*P(n, x)/(1 - x)^{2*n+1}, for n >= 0. This is proved with Lagrange's theorem applied to the Riordan triangle A135278 = (1/(1 - x)^2, x/(1 - x)). See an example below. - Wolfdieter Lang, Jul 31 2017 T(n,k) is the number of Dyck paths of semilength n with k-1 uu-blocks (pairs of consecutive up-steps). - Alexander Burstein, Jun 22 2020 In case you were searching for Narayama numbers, the correct spelling is Narayana. - N. J. A. Sloane, Nov 11 2020 Named after the Canadian mathematician Tadepalli Venkata Narayana (1930-1987). They were also called "Runyon numbers" after John P. Runyon (1922-2013) of Bell Telephone Laboratories, who used them in a study of a telephone traffic system. - Amiram Eldar, Apr 15 2021 The Narayana numbers were first studied by Percy Alexander MacMahon (see reference, Article 495) as pointed out by Bóna and Sagan (see link). - Peter Luschny, Apr 28 2022 T(n,k) is the degree distribution of the paths towards synchronization in the transition diagram associated with the Laplacian system over the complete graph K_n, corresponding to ordered initial conditions x_1 < x_2 < ... < x_n.
T(n,k) for n=2N+1 and k=N+1 is the number of states in the transition diagram associated with the Laplacian system over the complete bipartite graph K_{N,N}, corresponding to ordered (x_1 < x_2 < ... < x_N and x_{N+1} < x_{N+2} < ... < x_{2N}) and balanced (Sum_{i=1..N} x_i/N = Sum_{i=N+1..2N} x_i/N) initial conditions. (End)
Also the number of unlabeled ordered rooted trees with n nodes and k leaves. See the link by Marko Riedel. For example, row n = 5 counts the following trees:
((((o)))) (((o))o) ((o)oo) (oooo)
(((o)o)) ((oo)o)
(((oo))) ((ooo))
((o)(o)) (o(o)o)
((o(o))) (o(oo))
(o((o))) (oo(o))
The unordered version is A055277. Leaves in standard ordered trees are counted by A358371. (End)
REFERENCES
Berman and Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), pp. 103-124.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 196.
P. A. MacMahon, Combinatory Analysis, Vols. 1 and 2, Cambridge University Press, 1915, 1916; reprinted by Chelsea, 1960, Sect. 495.
T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, pp. 100-101.
A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.
T. K. Petersen, Eulerian Numbers, Birkhäuser, 2015, Chapter 2.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 17.
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.36(a) and (b).
LINKS
Axel Bacher, Antonio Bernini, Luca Ferrari, Benjamin Gunby, Renzo Pinzani, and Julian West, The Dyck pattern poset, Discrete Math. 321 (2014), 12--23. MR3154009. Aubrey Blecher, Charlotte Brennan, Arnold Knopfmacher, and Toufik Mansour, The perimeter of words, Discrete Mathematics, 340, no. 10 (2017): 2456-2465. R. Cori and G. Hetyei, How to count genus one partitions, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344. G. Kreweras, Sur les éventails de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #15 (1970), 3-41. [Annotated scanned copy] G. Kreweras, Sur les éventails de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #15 (1970), 3-41. Vincent Pilaud and V. Pons, Permutrees, arXiv:1606.09643 [math.CO], 2016-2017. Tad White, Quota Trees, arXiv:2401.01462 [math.CO], 2024. See pp. 19-20.
FORMULA
a(n, k) = C(n-1, k-1)*C(n, k-1)/k for k!=0; a(n, 0)=0.
Triangle equals [0, 1, 0, 1, 0, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is Deléham's operator defined in A084938. 0<n, 1<=k<=n a(n, 1) = a(n, n) = 1 a(n, k) = sum(i=1..n-1, sum(r=1..k-1, a(n-1-i, k-r) a(i, r))) + a(n-1, k) a(n, k) = sum(i=1..k-1, binomial(n+i-1, 2k-2)*a(k-1, i)) - Mike Zabrocki, Aug 26 2004 T(n, k) = C(n, k)*C(n-1, k-1) - C(n, k-1)*C(n-1, k) (determinant of a 2 X 2 subarray of Pascal's triangle A007318). - Gerald McGarvey, Feb 24 2005 T(n, k) = binomial(n-1, k-1)^2 - binomial(n-1, k)*binomial(n-1, k-2). - David Callan, Nov 02 2005 a(n,k) = C(n,2) (a(n-1,k)/((n-k)*(n-k+1)) + a(n-1,k-1)/(k*(k-1))) a(n,k) = C(n,k)*C(n,k-1)/n. - Mitch Harris, Jul 06 2006 G.f.: (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) = Sum_{n>0, k>0} a(n, k)*x^n*y^k.
Relation with Jacobi polynomials of parameter (1,1):
Row n+1 generating polynomial equals 1/(n+1)*x*(1-x)^n*Jacobi_P(n,1,1,(1+x)/(1-x)). It follows that the zeros of the Narayana polynomials are all real and nonpositive, as noted above. O.g.f for column k+2: 1/(k+1) * y^(k+2)/(1-y)^(k+3) * Jacobi_P(k,1,1,(1+y)/(1-y)). Cf. A008459. T(n+1,k) is the number of walks of n unit steps on the square lattice (i.e., each step in the direction either up (U), down (D), right (R) or left (L)) starting from the origin and finishing at lattice points on the x axis and which remain in the upper half-plane y >= 0 [Guy]. For example, T(4,3) = 6 counts the six walks RRL, LRR, RLR, UDL, URD and RUD, from the origin to the lattice point (1,0), each of 3 steps. Compare with tables A145596 - A145599. Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
The o.g.f. for this array is I(1 + t*x + t*x^2 + t*x^3 + ...) = 1 + t*x + (t + t^2)*x^2 + (t + 3*t^2 + t^3)*x^3 + ... = 1/(1 - x*t/(1 - x/(1 - x*t/(1 - x/(1 - ...))))) (as a continued fraction). Cf. A108767, A132081 and A141618. (End) G.f.: 1/(1-x-xy-x^2y/(1-x-xy-x^2y/(1-... (continued fraction). - Paul Barry, Sep 28 2010 E.g.f.: exp((1+y)x)*Bessel_I(1,2*sqrt(y)x)/(sqrt(y)*x). - Paul Barry, Sep 28 2010 G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n ). - Paul D. Hanna, Oct 13 2010 With F(x,t) = (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2))/(2*x) an o.g.f. in x for the Narayana polynomials in t, G(x,t) = x/(t+(1+t)*x+x^2) is the compositional inverse in x. Consequently, with H(x,t) = 1/ (dG(x,t)/dx) = (t+(1+t)*x+x^2)^2 / (t-x^2), the n-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*D_x)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*D_u)u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 04 2011 With offset 0, A001263 = Sum_{j>=0} A132710^j / A010790(j), a normalized Bessel fct. May be represented as the Pascal matrix A007318, n!/[(n-k)!*k!], umbralized with b(n)= A002378(n) for n>0 and b(0)=1: A001263(n,k)= b.(n!)/{b.[(n-k)!]*b.(k!)} where b.(n!) = b(n)*b(n-1)...*b(0), a generalized factorial (see example). - Tom Copeland, Sep 21 2011 With F(x,t) = {1-(1-t)*x-sqrt[1-2*(1+t)*x+[(t-1)*x]^2]}/2 a shifted o.g.f. in x for the Narayana polynomials in t, G(x,t)= x/[t-1+1/(1-x)] is the compositional inverse in x. Therefore, with H(x,t)=1/(dG(x,t)/dx)=[t-1+1/(1-x)]^2/{t-[x/(1-x)]^2}, (see A119900), the (n-1)-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*d/dx)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*d/du) u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 30 2011 T(n,k) = binomial(n-1,k-1)*binomial(n+1,k)-binomial(n,k-1)*binomial(n,k). - Philippe Deléham, Nov 05 2011 Damped sum of a column, in leading order: lim_{d->0} d^(2k-1) Sum_{N>=k} T(N,k)(1-d)^N=Catalan(n). - Joachim Wuttke, Sep 11 2014 Multiplying the n-th column by n! generates the revert of the unsigned Lah numbers, A089231. - Tom Copeland, Jan 07 2016 Row polynomials: (x - 1)^(n+1)*(P(n+1,(1 + x)/(x - 1)) - P(n-1,(1 + x)/(x - 1)))/((4*n + 2)), n = 1,2,... and where P(n,x) denotes the n-th Legendre polynomial. - Peter Bala, Mar 03 2017 The coefficients of the row polynomials R(n, x) = hypergeom([-n,-n-1], [2], x) generate the triangle based in (0,0). - Peter Luschny, Mar 19 2018 Multiplying the n-th diagonal by n!, with the main diagonal n=1, generates the Lah matrix A105278. With G equal to the infinitesimal generator of A132710, the Narayana triangle equals Sum_{n >= 0} G^n/((n+1)!*n!) = (sqrt(G))^(-1) * I_1(2*sqrt(G)), where G^0 is the identity matrix and I_1(x) is the modified Bessel function of the first kind of order 1. (cf. Sep 21 2011 formula also.) - Tom Copeland, Sep 23 2020 T(n,k) = T(n,k-1)*C(n-k+2,2)/C(k,2). - Yuchun Ji, Dec 21 2020
EXAMPLE
The initial rows of the triangle are:
[1] 1
[2] 1, 1
[3] 1, 3, 1
[4] 1, 6, 6, 1
[5] 1, 10, 20, 10, 1
[6] 1, 15, 50, 50, 15, 1
[7] 1, 21, 105, 175, 105, 21, 1
[8] 1, 28, 196, 490, 490, 196, 28, 1
[9] 1, 36, 336, 1176, 1764, 1176, 336, 36, 1;
...
For all n, 12...n (1 block) and 1|2|3|...|n (n blocks) are noncrossing set partitions.
Example of umbral representation:
A007318(5,k)=[1,5/1,5*4/(2*1),...,1]=(1,5,10,10,5,1),
so A001263(5,k)={1,b(5)/b(1),b(5)*b(4)/[b(2)*b(1)],...,1} = [1,30/2,30*20/(6*2),...,1]=(1,15,50,50,15,1).
First = last term = b.(5!)/[b.(0!)*b.(5!)]= 1. - Tom Copeland, Sep 21 2011 Row polynomials and diagonal sequences of A103371: n = 4, P(4, x) = 1 + 6*x + 6*x^2 + x^3, and the o.g.f. of fifth diagonal is G(4, x) = 5* P(4, x)/(1 - x)^9, namely [5, 75, 525, ...]. See a comment above. - Wolfdieter Lang, Jul 31 2017
MAPLE
A001263 := (n, k)->binomial(n-1, k-1)*binomial(n, k-1)/k;
a:=proc(n, k) option remember; local i; if k=1 or k=n then 1 else add(binomial(n+i-1, 2*k-2)*a(k-1, i), i=1..k-1); fi; end:
# Alternatively, as a (0, 0)-based triangle:
R := n -> simplify(hypergeom([-n, -n-1], [2], x)): Trow := n -> seq(coeff(R(n, x), x, j), j=0..n): seq(Trow(n), n=0..9); # Peter Luschny, Mar 19 2018
MATHEMATICA
T[n_, k_] := If[k==0, 0, Binomial[n-1, k-1] Binomial[n, k-1] / k];
Flatten[Table[Binomial[n-1, k-1] Binomial[n, k-1]/k, {n, 15}, {k, n}]] (* Harvey P. Dale, Feb 29 2012 *) TRow[n_] := CoefficientList[Hypergeometric2F1[1 - n, -n, 2, x], x];
Table[TRow[n], {n, 1, 11}] // Flatten (* Peter Luschny, Mar 19 2018 *) aot[n_]:=If[n==1, {{}}, Join@@Table[Tuples[aot/@c], {c, Join@@Permutations/@IntegerPartitions[n-1]}]];
Table[Length[Select[aot[n], Length[Position[#, {}]]==k&]], {n, 2, 9}, {k, 1, n-1}] (* Gus Wiseman, Jan 23 2023 *)
PROG
(PARI) {a(n, k) = if(k==0, 0, binomial(n-1, k-1) * binomial(n, k-1) / k)};
(PARI) {T(n, k)=polcoeff(polcoeff(exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*y^j)*x^m/m) +O(x^(n+1))), n, x), k, y)} \\ Paul D. Hanna, Oct 13 2010 (Haskell)
a001263 n k = a001263_tabl !! (n-1) !! (k-1)
a001263_row n = a001263_tabl !! (n-1)
a001263_tabl = zipWith dt a007318_tabl (tail a007318_tabl) where
dt us vs = zipWith (-) (zipWith (*) us (tail vs))
(zipWith (*) (tail us ++ [0]) (init vs))
(Magma) /* triangle */ [[Binomial(n-1, k-1)*Binomial(n, k-1)/k : k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 19 2014 (Sage)
@CachedFunction
def T(n, k):
if k == n or k == 1: return 1
if k <= 0 or k > n: return 0
return binomial(n, 2) * (T(n-1, k)/((n-k)*(n-k+1)) + T(n-1, k-1)/(k*(k-1)))
for n in (1..9): print([T(n, k) for k in (1..n)]) # Peter Luschny, Oct 28 2014 (GAP) Flat(List([1..11], n->List([1..n], k->Binomial(n-1, k-1)*Binomial(n, k-1)/k))); # Muniru A Asiru, Jul 12 2018
CROSSREFS
Row sums give A000108 (Catalan numbers), n>0. Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.
EXTENSIONS
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021
Number of binary words of length n (beginning with 0) whose autocorrelation function is the indicator of a singleton.
+10
29
1, 1, 2, 3, 6, 10, 20, 37, 74, 142, 284, 558, 1116, 2212, 4424, 8811, 17622, 35170, 70340, 140538, 281076, 561868, 1123736, 2246914, 4493828, 8986540, 17973080, 35943948, 71887896, 143771368, 287542736, 575076661, 1150153322, 2300289022, 4600578044, 9201120918
COMMENTS
The number of binary strings sharing the same autocorrelations.
The number of binary words of length n (beginning with 0) which do not start with an even palindrome (i.e. which are not of the form ss*t where s is a (nonempty) word, s* is its reverse, and t is any (possibly empty) word). - Mamuka Jibladze, Sep 30 2014 This sequence counts each of the following essentially equivalent things:
1. Sets of distinct positive integers with maximum n in which all adjacent elements have quotients > 1/2. For example, the a(1) = 1 through a(6) = 10 sets are:
{1} {2} {3} {4} {5} {6}
{2,3} {3,4} {3,5} {4,6}
{2,3,4} {4,5} {5,6}
{2,3,5} {3,4,6}
{3,4,5} {3,5,6}
{2,3,4,5} {4,5,6}
{2,3,4,6}
{2,3,5,6}
{3,4,5,6}
{2,3,4,5,6}
2. For n > 1, sets of distinct positive integers with maximum n - 1 whose first-differences are term-wise less than their decapitation (remove the maximum). For example, the set q = {2,4,5} has first-differences (2,1), which are not less than (2,4), so q is not counted under a(5). On the other hand, r = {2,3,5,6} has first-differences {1,2,1}, which are less than {2,3,5}, so r is counted under a(6).
3. Compositions of n where each part after the first is less than the sum of all preceding parts. For example, the a(1) = 1 through a(6) = 10 compositions are:
(1) (2) (3) (4) (5) (6)
(21) (31) (41) (51)
(211) (32) (42)
(311) (411)
(212) (321)
(2111) (312)
(3111)
(2121)
(2112)
(21111)
(End)
FORMULA
a(2n) = 2*a(2n-1) - a(n) for n >= 1; a(2n+1) = 2*a(2n) for n >= 1.
MAPLE
a:= proc(n) option remember; `if`(n=0, 1/2,
2*a(n-1)-`if`(n::odd, 0, a(n/2)))
end:
MATHEMATICA
a[1] = 1; a[n_] := a[n] = If[EvenQ[n], 2*a[n-1] - a[n/2], 2*a[n-1]]; Array[a, 40] (* Jean-François Alcover, Jul 17 2015 *) Table[Length[Select[Subsets[Range[n]], MemberQ[#, n]&&Min@@Divide@@@Partition[#, 2, 1]>1/2&]], {n, 8}] (* Gus Wiseman, Mar 08 2021 *)
PROG
(PARI) a(n)=if(n<2, n>0, 2*a(n-1)-(1-n%2)*a(n\2))
CROSSREFS
The version with quotients <= 1/2 is A018819. The version with quotients < 1/2 is A040039. A000045 counts sets containing n with all differences > 2.
A000929 counts partitions with no adjacent parts having quotient < 2.
A342094 counts partitions with no adjacent parts having quotient > 2.
AUTHOR
Torsten.Sillke(AT)uni-bielefeld.de
Atkinson-Negro-Santoro sequence: a(n+1) = 2*a(n) - a(n-floor(n/2+1)).
(Formerly M1076)
+10
7
0, 1, 2, 4, 7, 13, 24, 46, 88, 172, 337, 667, 1321, 2629, 5234, 10444, 20842, 41638, 83188, 166288, 332404, 664636, 1328935, 2657533, 5314399, 10628131, 21254941, 42508561, 85014493, 170026357, 340047480, 680089726, 1360169008, 2720327572
COMMENTS
For each n, the n-term sequence (b(k) = a(n) - a(n-k), 1 <= k <= n), has the property that all 2^n sums of subsets of the terms are distinct.
REFERENCES
S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.28.
T. V. Narayana, Recent progress and unsolved problems in dominance theory, pp. 68-78 of Combinatorial mathematics (Canberra 1977), Lect. Notes Math. Vol. 686, 1978.
T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, pp. 100-101.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
EXAMPLE
For n = 4, the sequence b is 7-4,7-2,7-1,7-0 = 3,5,6,7, which has subset sums (grouped by number of terms) 0, 3,5,6,7, 8,9,10,11,12,13, 14,15,16,18, 21.
MATHEMATICA
a[ 0 ] := 0; a[ 1 ] := 1; a[ n_ ] := 2*a[ n - 1 ] - a[(n - 1) - Floor[ (n - 1)/2 + 1 ] ]; For[ n = 1, n <= 100, n++, Print[ a[ n ] ] ];
PROG
(Haskell)
a005255 n = a005255_list !! (n-1)
a005255_list = scanl (+) 0 $ tail a002083_list
EXTENSIONS
More terms from Winston C. Yang (winston(AT)cs.wisc.edu), Aug 26 2000
a(n) is the number of permutations avoiding 231 and 312 realizable on increasing strict binary trees with 2n-1 nodes.
+10
6
1, 2, 6, 22, 84, 330, 1308, 5210, 20796, 83100, 332232, 1328598, 5313732, 21253620, 85011864, 340042246, 1360158564
COMMENTS
The number of permutations avoiding 231 and 312 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. A strict binary tree is a tree graph where each node has 0 or 2 children. The permutation is found by reading the labels in the order they appear in a breadth-first search. (Note that breadth-first search reading word is equivalent to reading the tree left to right by levels, starting with the root.)
In some cases, more than one tree results in the same breadth-first search reading word, but here we count the permutations, not the trees.
EXAMPLE
For example, when n=3, the permutations 12543, 12435, 13245, 13254, 12345,and 12354. all avoid 231 and 312 in the classical sense and occur as breadth-first search reading words on an increasing strict binary tree with 5 nodes.
. 1 1 1 1 1 1
. / \ / \ / \ / \ / \ / \
. 2 5 2 4 3 2 3 2 2 3 2 3
. / \ / \ / \ / \ / \ / \
. 4 3 3 5 4 5 5 4 4 5 5 4
Triangle read by rows: T(n,k) (n >= 1, 1 <= k< = n) gives number of non-distorting tie-avoiding integer vote weights.
+10
5
1, 1, 2, 2, 3, 4, 3, 5, 6, 7, 6, 9, 11, 12, 13, 11, 17, 20, 22, 23, 24, 22, 33, 39, 42, 44, 45, 46, 42, 64, 75, 81, 84, 86, 87, 88, 84, 126, 148, 159, 165, 168, 170, 171, 172, 165, 249, 291, 313, 324, 330, 333, 335, 336, 337, 330, 495, 579, 621, 643, 654, 660, 663, 665
REFERENCES
Author?, Solution to Board of Directors Problem, J. Rec. Math., 9 (No. 3, 1977), 240.
T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, pp. 100-101.
FORMULA
T(1,1) = 1;
T(n,1) = T(n-1, floor((n+1)/2));
T(n,k) = T(n,1) + T(n-1,k-1) for k > 1.
EXAMPLE
Triangle:
1;
1, 2;
2, 3, 4;
3, 5, 6, 7;
6, 9, 11, 12, 13;
...
MATHEMATICA
a[1, 1] = 1; a[n_, 1] := a[n, 1] = a[n - 1, Floor[(n + 1)/2]]; a[n_, k_ /; k > 1] := a[n, k] = a[n, 1] + a[n - 1, k - 1]; A037254 = Flatten[ Table[ a[n, k], {n, 1, 11}, {k, 1, n}]] (* Jean-François Alcover, Apr 03 2012, after given recurrence *)
PROG
(Haskell)
a037254 n k = a037254_tabl !! (n-1) !! (k-1)
a037254_row n = a037254_tabl !! (n-1)
a037254_tabl = map fst $ iterate f ([1], drop 2 a002083_list) where
f (row, (x:xs)) = (map (+ x) (0 : row), xs)
(Python)
def T(n, k):
if k==1:
if n==1: return 1
else: return T(n - 1, (n + 1)//2)
return T(n, 1) + T(n - 1, k - 1)
for n in range(1, 12): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Jun 03 2017
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