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Multiplication table read by antidiagonals: T(i,j) = i*j, i>=1, j>=1.
+10
113
1, 2, 2, 3, 4, 3, 4, 6, 6, 4, 5, 8, 9, 8, 5, 6, 10, 12, 12, 10, 6, 7, 12, 15, 16, 15, 12, 7, 8, 14, 18, 20, 20, 18, 14, 8, 9, 16, 21, 24, 25, 24, 21, 16, 9, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 11, 20, 27, 32, 35, 36, 35, 32, 27, 20, 11, 12, 22, 30, 36, 40, 42, 42, 40, 36, 30, 22, 12
COMMENTS
Or, triangle X(n,m) = T(n-m+1,m) read by rows, in which row n gives the numbers n*1, (n-1)*2, (n-2)*3, ..., 2*(n-1), 1*n.
Radius of incircle of Pythagorean triangle with sides a=(n+1)^2-m^2, b=2*(n+1)*m and c=(n+1)^2+m^2. - Floor van Lamoen, Aug 16 2001 In the proof of countability of rational numbers they are arranged in a square array. a(n) = p*q where p/q is the corresponding rational number as read from the array. - Amarnath Murthy, May 29 2003 Row 12 gives total number of partridges, turtle doves, ... and drummers drumming that you have received at the end of the Twelve Days of Christmas song. - Alonso del Arte, Jun 17 2005 Consider a particle with spin S (a half-integer) and 2S+1 quantum states |m>, m = -S,-S+1,...,S-1,S. Then the matrix element <m+1|S_+|m> = sqrt((S+m+1)(S-m)) of the spin-raising operator is the square-root of the triangular (tabl) element T(r,o) of this sequence in row r = 2S, and at offset o=2(S+m). T(r,o) is also the intensity |<m+1|S_+|m><m|S_-|m+1>| of the transition between the states |m> and |m+1>. For example, the five transitions between the 6 states of a spin S=5/2 particle have relative intensities 5,8,9,8,5. The total intensity of all spin 5/2 transitions (relative to spin 1/2) is 35, which is the tetrahedral number A000292(5). - Stanislav Sykora, May 26 2012 T(n, k) is also the (k-1)-superdiagonal sum of an n X n Toeplitz matrix M(n) whose first row consists of successive positive integer numbers 1, ..., n. - Stefano Spezia, Jul 12 2019 X(n, m+1) is the number of degrees of freedom that an m-dimensional flat geometry (point, line, plane, etc.) has when embedded in an n-dimensional Euclidean space.
X(n+1, m+1) is the number of degrees of freedom that an m-ball has when embedded in an n-dimensional Euclidean space. (End)
REFERENCES
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 46.
FORMULA
Rectangular array: T(n, m) = n*m, n>=1, m>= 1.
Triangle X(n, m) = T(n-m+1, m) = (n-m+1)*m.
Sum_{i=1..n} Sum_{j=1..n} a(n) = A000537(n) [Sum of first n cubes; or n-th triangular number squared.] Determinant of all n X n contiguous subarrays of A003991 is 0. - Gerald McGarvey, Sep 26 2004 G.f. as rectangular array: x * y / [ (1-x)^2 * (1-y)^2 ].
a(n) = i*j, where i=floor((1+sqrt(8n-7))/2), j=n-i*(i-1)/2. - Hieronymus Fischer, Aug 08 2007 As a linear array, the sequence is a(n) = A002260(n)* A004736(n) or a(n) = ((t*t+3*t+4)/2-n)*(n-(t*(t+1)/2)), where t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 17 2012 G.f. as linear array: (x - 3*x^2 + Sum_{k >= 0} ((k+2-x-(k+1)*x^2)*x^((k^2+3*k+4)/2)))/(1-x)^3. - Robert Israel, Dec 14 2015 E.g.f. as triangle: exp(x+y)*(1 + x - y + x*y - y^2). - Stefano Spezia, Jul 12 2019 a(n) = (1/2)*t + (n - 1/4)*t^2 - (1/4)*t^4 - n^2 + n, where t = floor(sqrt(2*n) + 1/2). - Ridouane Oudra, Nov 21 2020 T(n,k) = number of sums |x-y|+|y-z| = k, where x,y,z are in {1,2,...,n} and x < y < z. - Clark Kimberling, Jan 22 2024
EXAMPLE
The array T starts in row n=1 with columns m>=1 as:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
3 6 9 12 15 18 21 24 27 30 33 36 39 42 45
4 8 12 16 20 24 28 32 36 40 44 48 52 56 60
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75
6 12 18 24 30 36 42 48 54 60 66 72 78 84 90
7 14 21 28 35 42 49 56 63 70 77 84 91 98 105
8 16 24 32 40 48 56 64 72 80 88 96 104 112 120
9 18 27 36 45 54 63 72 81 90 99 108 117 126 135
10 20 30 40 50 60 70 80 90 100 110 120 130 140 150
The triangle X(n, m) begins
n\m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ...
1: 1
2: 2 2
3: 3 4 3
4: 4 6 6 4
5: 5 8 9 8 5
6: 6 10 12 12 10 6
7: 7 12 15 16 15 12 7
8: 8 14 18 20 20 18 14 8
9: 9 16 21 24 25 24 21 16 9
10: 10 18 24 28 30 30 28 24 18 10
11: 11 20 27 32 35 36 35 32 27 20 11
12: 12 22 30 36 40 42 42 40 36 30 22 12
13: 13 24 33 40 45 48 49 48 45 40 33 24 13
14: 14 26 36 44 50 54 56 56 54 50 44 36 26 14
15: 15 28 39 48 55 60 63 64 63 60 55 48 39 28 15
MAPLE
seq(seq(i*(n-i), i=1..n-1), n=2..10); # Robert Israel, Dec 14 2015
MATHEMATICA
f[n_] := Table[SeriesCoefficient[E^(x + y) (1+ x - y +x*y-y^2), {x, 0, i}, {y, 0, j}]*i!*j!, {i, n, n}, {j, 0, n}]; Flatten[Array[f, 11, 0]] (* Stefano Spezia, Jul 12 2019 *)
PROG
(PARI) A003991(n, k) = if(k<1 || n<1, 0, k*n) (Magma) /* As triangle */ [[k*(n-k+1): k in [1..n]]: n in [1..15]]; // Vincenzo Librandi, Jul 12 2019
CROSSREFS
Main diagonal gives squares A000290. Antidiagonal sums are tetrahedral numbers A000292. See A004247 for another version. Cf. A003989, A003990, A003056, A049581, A000442, A027424, A002260, A033638, A059895, A059896, A059897, A002620.
Numbers of the form 2^i - 2^j with i >= j.
+10
72
0, 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 28, 30, 31, 32, 48, 56, 60, 62, 63, 64, 96, 112, 120, 124, 126, 127, 128, 192, 224, 240, 248, 252, 254, 255, 256, 384, 448, 480, 496, 504, 508, 510, 511, 512, 768, 896, 960, 992, 1008, 1016, 1020, 1022, 1023
COMMENTS
Numbers whose digits in base 2 are in nonincreasing order.
Might be called "nialpdromes".
Subset of A077436. Proof: Since a(n) is of the form (2^i-1)*2^j, i,j >= 0, a(n)^2 = (2^(2i) - 2^(i+1))*2^(2j) + 2^(2j) where the first sum term has i-1 one bits and its 2j-th bit is zero, while the second sum term switches the 2j-th bit to one, giving i one bits, as in a(n). - Ralf Stephan, Mar 08 2004 Numbers whose binary representation contains no "01". - Benoit Cloitre, May 23 2004 Every polynomial with coefficients equal to 1 for the leading terms and 0 after that, evaluated at 2. For instance a(13) = x^4 + x^3 + x^2 at 2, a(14) = x^4 + x^3 + x^2 + x at 2. - Ben Paul Thurston, Jan 11 2008 As a triangle by rows starting:
1;
2, 3;
4, 6, 7;
8, 12, 14, 15;
16, 24, 28, 30, 31;
...,
equals A000012 * A130123 * A000012, where A130123 = (1, 0,2; 0,0,4; 0,0,0,8; ...). Row sums of this triangle = A000337 starting (1, 5, 17, 49, 129, ...). (End) First differences are A057728 = 1; 1; 1; 1; 2,1; 1; 4,2,1; 1; 8,4,2,1; 1; ... i.e., decreasing powers of 2, separated by another "1". - M. F. Hasler, May 06 2009 Apart from first term, numbers that are powers of 2 or the sum of some consecutive powers of 2. - Omar E. Pol, Feb 14 2013 Numbers that can be digitally generated with twisted ring (Johnson) counters. This is, the binary digits of a(n) correspond to those stored in a shift register where the input bit of the first bit storage element is the inverted output of the last storage element. After starting with all 0’s, each new state is obtained by rotating the stored bits but inverting at each state transition the last bit that goes to the first position (see link).
Examples: for a(n) represented by three bits
Binary
a(5)= 4 -> 100 last bit = 0
a(6)= 6 -> 110 first bit = 1 (inverted last bit of previous number)
a(7)= 7 -> 111
and for a(n) represented by four bits
Binary
a(8) = 8 -> 1000
a(9) = 12 -> 1100 last bit = 0
a(10)= 14 -> 1110 first bit = 1 (inverted last bit of previous number)
a(11)= 15 -> 1111
(End)
Powers of 2 represented in bases which are terms of this sequence must always contain at least one digit which is also a power of 2. This is because 2^i mod (2^i - 2^j) = 2^j, which means the last digit always cycles through powers of 2 (or if i=j+1 then the first digit is a power of 2 and the rest are trailing zeros). The only known non-member of this sequence with this property is 5. - Ely Golden, Sep 05 2017
LINKS
Eric Weisstein's World of Mathematics, Digit.
FORMULA
a(n) = 2^s(n) - 2^((s(n)^2 + s(n) - 2n)/2) where s(n) = ceiling((-1 + sqrt(1+8n))/2). - Sam Alexander, Jan 08 2005 a(n) = 2^k + a(n-k-1) for 1 < n and k = A003056(n-2). The rows of T(r, c) = 2^r-2^c for 0 <= c < r read from right to left produce this sequence: 1; 2, 3; 4, 6, 7; 8, 12, 14, 15; ... - Frank Ellermann, Dec 06 2001 a(n+1) = (2^(n - r(r-1)/2) - 1) 2^(r(r+1)/2 - n), where r=round(sqrt(2n)). - M. F. Hasler, May 06 2009 G.f.: (x^2/((2-x)*(1-x)))*(1 + Sum_{k>=0} x^((k^2+k)/2)*(1 + x*(2^k-1))). The sum is related to Jacobi theta functions. - Robert Israel, Feb 24 2015 a(n) = a(n-1) + a(n-d)/a(d*(d+1)/2 + 2) if n > 1, d > 0, where d = A002262(n-2). - Yuchun Ji, May 11 2020
EXAMPLE
a(22) = 64 = 32 + 32 = 2^5 + a(16) = 2^ A003056(20) + a(22-5-1). a(23) = 96 = 64 + 32 = 2^6 + a(16) = 2^ A003056(21) + a(23-6-1). a(24) = 112 = 64 + 48 = 2^6 + a(17) = 2^ A003056(22) + a(24-6-1).
MAPLE
a:=proc(n) local n2, d: n2:=convert(n, base, 2): d:={seq(n2[j]-n2[j-1], j=2..nops(n2))}: if n=0 then 0 elif n=1 then 1 elif d={0, 1} or d={0} or d={1} then n else fi end: seq(a(n), n=0..2100); # Emeric Deutsch, Apr 22 2006
MATHEMATICA
Union[Flatten[Table[2^i - 2^j, {i, 0, 100}, {j, 0, i}]]] (* T. D. Noe, Mar 15 2011 *) Select[Range[0, 2^10], NoneTrue[Differences@ IntegerDigits[#, 2], # > 0 &] &] (* Michael De Vlieger, Sep 05 2017 *)
PROG
(PARI) for(n=0, 2500, if(prod(k=1, length(binary(n))-1, component(binary(n), k)+1-component(binary(n), k+1))>0, print1(n, ", ")))
(PARI) A023758(n)= my(r=round(sqrt(2*n--))); (1<<(n-r*(r-1)/2)-1)<<(r*(r+1)/2-n) /* or, to illustrate the "decreasing digit" property and analogy to A064222: */ A023758(n, show=0)={ my(a=0); while(n--, show & print1(a", "); a=vecsort(binary(a+1)); a*=vector(#a, j, 2^(j-1))~); a} \\ M. F. Hasler, May 06 2009
(PARI) list(lim)=my(v=List([0]), t); for(i=1, logint(lim\1+1, 2), t=2^i-1; while(t<=lim, listput(v, t); t*=2)); Set(v) \\ Charles R Greathouse IV, May 03 2016 (Haskell)
import Data.Set (singleton, deleteFindMin, insert)
a023758 n = a023758_list !! (n-1)
a023758_list = 0 : f (singleton 1) where
f s = x : f (if even x then insert z s' else insert z $ insert (z+1) s')
where z = 2*x; (x, s') = deleteFindMin s
(Python)
def a_next(a_n): return (a_n | (a_n >> 1)) + (a_n & 1)
a_n = 1; a = [0]
for i in range(55): a.append(a_n); a_n = a_next(a_n) # Falk Hüffner, Feb 19 2022
CROSSREFS
Positions of nonzero terms in A341509 (apart from the initial zero). Positions of squarefree terms in A260443.
Number of middle divisors of n, i.e., divisors in the half-open interval [sqrt(n/2), sqrt(n*2)).
+10
72
1, 1, 0, 1, 0, 2, 0, 1, 1, 0, 0, 2, 0, 0, 2, 1, 0, 1, 0, 2, 0, 0, 0, 2, 1, 0, 0, 2, 0, 2, 0, 1, 0, 0, 2, 1, 0, 0, 0, 2, 0, 2, 0, 0, 2, 0, 0, 2, 1, 1, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 0, 2, 1, 0, 2, 0, 0, 0, 2, 0, 3, 0, 0, 0, 0, 2, 0, 0, 2, 1, 0, 0, 2, 0, 0, 0, 2, 0, 2, 2, 0, 0, 0, 0, 2, 0, 1, 2, 1, 0, 0, 0, 2, 0
COMMENTS
Theorem 1: (i) a(n) = (number of odd divisors of n <= sqrt(2*n)) - (number of odd divisors of n > sqrt(2*n)).
(ii) Let r(n) = A003056(n). Then a(n) = (number of odd divisors of n <= r(n)) - (number of odd divisors of n > r(n)). (iii) a(n) = Sum_{k=1..r(n)} (-1)^(k+1)* A237048(n,k). (iv) a(n) is odd iff n is a square or twice a square (cf. A053866). Indices of odd terms give A028982. Indices of even terms give A028983. The proofs are straightforward. These results were conjectured by Omar E. Pol in 2017. (End) Theorem 2: a(n) is equal to the difference between the number of partitions of n into an odd number of consecutive parts and the number of partitions of n into an even number of consecutive parts. [Chapman et al., 2001; Hirschhorn and Hirschhorn, 2005]. - Omar E. Pol, Feb 06 2017 Conjecture 1: This is the central column of the isosceles triangle of A249351. Conjecture 2: a(n) is also the width of the terrace at the n-th level in the main diagonal of the pyramid described in A245092. Conjecture 3: a(n) is also the number of central subparts of the symmetric representation of sigma(n). For more information see A279387.
REFERENCES
Robin Chapman, Kimmo Eriksson and Richard Stanley, On the Number of Divisors of n in a Special Interval: Problem 10847, Amer. Math. Monthly 108:1 (Jan 2001), p. 77 (Proposal); 109:1 (Jan 2002), p. 80 (Solution). [Please do not delete this reference. - N. J. A. Sloane, Dec 16 2020]
FORMULA
G.f.: Sum_{k>=1} (-1)^(k-1)*x^(k*(k+1)/2)/(1-x^k). (This g.f. corresponds to the assertion in Theorem 2.)
Another g.f., corresponding to the definition: Sum_{k>=1} x^(2*k*(k+1))*(1-x^(6*k^2))/(1-x^(2*k)) + Sum_{k>=0} x^((k+1)*(2*k+1))*(1-x^((2*k+1)*(3*k+2))/(1-x^(2*k+1). - N. J. A. Sloane, Jan 04 2021
EXAMPLE
a(6)=2 because the 2 middle divisors of 6 (2 and 3) are between sqrt(3) and sqrt(12).
MATHEMATICA
(* number of middle divisors *)
a067742[n_] := Select[Divisors[n], Sqrt[n/2] <= # < Sqrt[2n] &]
a067742[115] (* data *)
a[ n_] := If[ n < 1, 0, DivisorSum[ n, 1 &, n/2 <= #^2 < 2 n &]]; (* Michael Somos, Jun 04 2015 *) (* support function a240542[] is defined in A240542 *) b[n_] := a240542[n] - a240542[n-1]
PROG
(PARI) A067742(n) = {sumdiv(n, d, d2 = d^2; n / 2 < d2 && d2 <= n << 1)} \\ M. F. Hasler, May 12 2008 (PARI) a(n) = A067742(n) = {my(d = divisors(n), iu = il = #d \ 2); if(#d <= 2, return(n < 3)); while(d[il]^2 > n>>1, il--); while(d[iu]^2 < (n<<1), iu++); (Python)
from sympy import divisors
def A067742(n): return sum(1 for d in divisors(n, generator=True) if n <= 2*d**2 < 4*n) # Chai Wah Wu, Jun 09 2022
CROSSREFS
Cf. also A071561 (positions of zeros), A071562 (positions of nonzeros), A299761 (middle divisors of n), A355143 (products of middle divisors).
Triangle read by rows in which row n lists two copies of the n-th row of triangle A237593.
+10
69
1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 2, 3, 1, 1, 3, 3, 1, 1, 3, 3, 2, 2, 3, 3, 2, 2, 3, 4, 1, 1, 1, 1, 4, 4, 1, 1, 1, 1, 4, 4, 2, 1, 1, 2, 4, 4, 2, 1, 1, 2, 4, 5, 2, 1, 1, 2, 5, 5, 2, 1, 1, 2, 5, 5, 2, 2, 2, 2, 5, 5, 2, 2, 2, 2, 5, 6, 2, 1, 1, 1, 1, 2, 6, 6, 2, 1, 1, 1, 1, 2, 6, 6, 3, 1, 1, 1, 1, 3, 6, 6, 3, 1, 1, 1, 1, 3, 6
COMMENTS
For the construction of this sequence also we can start from A235791. This sequence can be interpreted as an infinite Dyck path: UDUDUUDD...
Also we use this sequence for the construction of a spiral in which the arms in the quadrants give the symmetric representation of sigma, see example.
We can find the spiral (mentioned above) on the terraces of the stepped pyramid described in A244050. - Omar E. Pol, Dec 07 2016 The spiral has the property that the sum of the parts in the quadrants 1 and 3, divided by the sum of the parts in the quadrants 2 and 4, converges to 3/5. - Omar E. Pol, Jun 10 2019
EXAMPLE
Triangle begins (first 15.5 rows):
1, 1, 1, 1;
2, 2, 2, 2;
2, 1, 1, 2, 2, 1, 1, 2;
3, 1, 1, 3, 3, 1, 1, 3;
3, 2, 2, 3, 3, 2, 2, 3;
4, 1, 1, 1, 1, 4, 4, 1, 1, 1, 1, 4;
4, 2, 1, 1, 2, 4, 4, 2, 1, 1, 2, 4;
5, 2, 1, 1, 2, 5, 5, 2, 1, 1, 2, 5;
5, 2, 2, 2, 2, 5, 5, 2, 2, 2, 2, 5;
6, 2, 1, 1, 1, 1, 2, 6, 6, 2, 1, 1, 1, 1, 2, 6;
6, 3, 1, 1, 1, 1, 3, 6, 6, 3, 1, 1, 1, 1, 3, 6;
7, 2, 2, 1, 1, 2, 2, 7, 7, 2, 2, 1, 1, 2, 2, 7;
7, 3, 2, 1, 1, 2, 3, 7, 7, 3, 2, 1, 1, 2, 3, 7;
8, 3, 1, 2, 2, 1, 3, 8, 8, 3, 1, 2, 2, 1, 3, 8;
8, 3, 2, 1, 1, 1, 1, 2, 3, 8, 8, 3, 2, 1, 1, 1, 1, 2, 3, 8;
9, 3, 2, 1, 1, 1, 1, 2, 3, 9, ...
.
Illustration of initial terms as an infinite Dyck path (row n = 1..4):
.
. /\/\ /\/\
. /\ /\ /\/\ /\/\ / \ / \
. /\/\/ \/ \/ \/ \/ \/ \
.
.
Illustration of initial terms for the construction of a spiral related to sigma:
.
. row 1 row 2 row 3 row 4
. _ _ _
. |_
. _ _ |
. _ _ | |
. | | | |
. | | | |
. |_ _ |_ _| |
. |_ _ _ _| _|
. _ _ _|
.
.[1,1,1,1] [2,2,2,2] [2,1,1,2,2,1,1,2] [3,1,1,3,3,1,1,3]
.
The first 2* A003056(n) terms of the n-th row are represented in the A010883(n-1) quadrant and the last 2* A003056(n) terms of the n-th row are represented in the A010883(n) quadrant. .
Illustration of the spiral constructed with the first 15.5 rows of triangle:
.
. 12 _ _ _ _ _ _ _ _
. | _ _ _ _ _ _ _|_ _ _ _ _ _ _ 7
. | | |_ _ _ _ _ _ _|
. _| | |
. |_ _|9 _ _ _ _ _ _ |_ _
. 12 _ _| | _ _ _ _ _|_ _ _ _ _ 5 |_
. _ _ _| | _| | |_ _ _ _ _| |
. | _ _ _| 9 _|_ _| |_ _ 3 |_ _ _ 7
. | | _ _| | 12 _ _ _ _ |_ | | |
. | | | _ _| _| _ _ _|_ _ _ 3 |_|_ _ 5 | |
. | | | | _| | |_ _ _| | | | |
. | | | | | _ _| |_ _ 3 | | | |
. | | | | | | 3 _ _ | | | | | |
. | | | | | | | _|_ 1 | | | | | |
. _|_| _|_| _|_| _|_| |_| _|_| _|_| _|_| _
. | | | | | | | | | | | | | | | |
. | | | | | | |_|_ _ _| | | | | | | |
. | | | | | | 2 |_ _|_ _| _| | | | | | |
. | | | | |_|_ 2 |_ _ _| _ _| | | | | |
. | | | | 4 |_ 7 _| _ _| | | | |
. | | |_|_ _ |_ _ _ _ | _| _ _ _| | | |
. | | 6 |_ |_ _ _ _|_ _ _ _| | _| _ _| | |
. |_|_ _ _ |_ 4 |_ _ _ _ _| _| | _ _ _| |
. 8 | |_ _ | 15| _| | _ _ _|
. |_ | |_ _ _ _ _ _ | _ _| _| |
. 8 |_ |_ |_ _ _ _ _ _|_ _ _ _ _ _| | _| _|
. |_ _| 6 |_ _ _ _ _ _ _| _ _| _|
. | 28| _ _|
. |_ _ _ _ _ _ _ _ | |
. |_ _ _ _ _ _ _ _|_ _ _ _ _ _ _ _| |
. 8 |_ _ _ _ _ _ _ _ _|
. 31
.
The diagram contains A237590(16) = 27 parts. The total area (also the total number of cells) in the n-th arm of the spiral is equal to sigma(n) = A000203(n), considering every quadrant and the axes x and y. (checked by hand up to row n = 128). The parts of the spiral are in A237270: 1, 3, 2, 2, 7...
CROSSREFS
The sum of row n is equal to 4*n = A008586(n). Row n is a palindromic composition of 4*n = A008586(n). Both column 1 and right border are A008619, n >= 1. Cf. A000203, A000217, A003056, A008619, A010883, A112610, A193553, A237048, A237271, A237590, A239052, A239053, A239663, A239665, A239931, A239932, A239933, A239934, A240020, A240062, A244050, A245092, A262626, A296508, A299778.
Triangle read by rows: row n gives partial alternating sums of row n of A237048.
+10
69
1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 2, 2, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 2, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 2, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 2, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 2, 2, 2, 2
COMMENTS
All entries in the triangle are nonnegative since the number of 1's in odd-numbered columns of A237048 prior to column j, 1 <= j <= row(n), is at least as large as the number of 1's in even-numbered columns through column j. As a consequence: (a) The two adjacent symmetric Dyck paths whose legs are defined by adjacent rows of triangle A237593 never cross each other (see also A235791 and A237591) and the rows in this triangle describe the widths between the legs. (b) Let legs(n) denote the n-th row of triangle A237591, widths(n) the n-th row of this triangle, and c(n) the rightmost entry in the n-th row of this triangle (center of the Dyck path). Then area(n) = 2 * legs(n) . width(n) - c(n), where "." is the inner product, is the area between the two adjacent symmetric Dyck paths. For a proof that T(n, k) = | { d : d|n and k/2 < d <= k }, for 1 <= k <= row(n), an identity suggested by Peter Munn, see the link. A corollary to it is that the number of divisors of n in the half-open interval (row(n)/2, row(n)] equals the width of the symmetric representation of n at the diagonal: T(n, row(n)) = | { d : d|n and row(n)/2 < d <= row(n) } |. See also the comments and conjectures of Michel Marcus in A067742 and A237593. - Hartmut F. W. Hoft, Jun 24 2024 Conjecture 1: Every column is a periodic sequence.
Conjecture 2: The periods of the columns 1..8 are respectively: 1, 2, 6, 12, 60, 60, 420, 840.
Question 1: Is the period of the column k equal to A003418(k)? (End). Conjecture 3: Column 3 gives [2, 0] together with A115357, hence column 3 gives 2 together with A171182. Question 2: Except the first nine terms of A337976, is the column 4 the same as A337976? Question 3: Except the first 14 terms of A366981, is the column 5 the same as A366981? (End) Conjectures 1 and 2 are true and the answer to question 1 is affirmative.
By definition, each column k in triangle T237048(n, k) of sequence A237048 is a periodic sequence of period k. Since the k-th term in row n of the triangle T(n, k) = Sum_{i=1 .. k) ( (-1)^(i+1) * T237048(n, i) ), with 1 <= k <= A003056(n), each initial subsequence T(n, 1) .. T(n, k) of row n in this triangle is periodic of period lcm(1, .. , k) = A003418(k). This implies that each column k in this sequence has period A003418(k). Conjecture 3 and Question 2 are true. Since T237048(n, 1) = 1, T237208(n, 2) = 1 if n odd and 0 if n even, T237048(n, 3) = 1 if 3|n and 0 otherwise, and T237048(n, 4) = 1 if 4|(n-2) and 0 otherwise, equations T249223(n, 3) = 1 - (n mod 2) + delta( n mod 3) and T249223(n, 4) = 1 - (n mod 2) + delta( n mod 3) - delta( (n-2) mod 4) hold where delta(k) = 1 if k = 0 and 0 otherwise. With the 3rd column starting at n = A000217(3) = 6, each period starting in a row that is a multiple of 6 is [ 2 0 1 1 1 0 ], and appropriate shifts yield A115357 and A171182. With the 4th column starting at n = A000217(4) = 10, each period starting in a row n with 12|(n+2) is [ 0 0 2 0 0 1 1 0 1 0 1 1 ], and with a shift of 9 yields the apparently periodic A337976(10), A337976(11), ... (End)
FORMULA
T(n, k) = Sum_{j=1..k} (-1)^(j+1)* A237048(n, j), for n>=1 and 1 <= k <= floor((sqrt(8*n + 1) - 1)/2). - corrected by Hartmut F. W. Hoft, Jan 25 2018
EXAMPLE
Triangle begins:
---------------------------
n \ k 1 2 3 4 5 6
---------------------------
1 | 1;
2 | 1;
3 | 1, 0;
4 | 1, 1;
5 | 1, 0;
6 | 1, 1, 2;
7 | 1, 0, 0;
8 | 1, 1, 1;
9 | 1, 0, 1;
10 | 1, 1, 1, 0;
11 | 1, 0, 0, 0;
12 | 1, 1, 2, 2;
13 | 1, 0, 0, 0;
14 | 1, 1, 1, 0;
15 | 1, 0, 1, 1, 2;
16 | 1, 1, 1, 1, 1;
17 | 1, 0, 0, 0, 0;
18 | 1, 1, 2, 1, 1;
19 | 1, 0, 0, 0, 0;
20 | 1, 1, 1, 1, 2;
21 | 1, 0, 1, 1, 1, 0;
22 | 1, 1, 1, 0, 0, 0;
23 | 1, 0, 0, 0, 0, 0;
24 | 1, 1, 2, 2, 2, 2;
...
The triangle shows that area(n) has width 1 for powers of 2 and that area(p) for primes p consists of only 1 horizontal leg of width 1 (and its symmetric vertical leg in the mirror symmetric duplicate of this triangle).
MAPLE
r := proc(n) floor((sqrt(1+8*n)-1)/2) ; end proc: # R. J. Mathar 2015 A003056A237048:=proc(n, k) local i; global r;
if n<(k-1)*k/2 or k>r(n) then return(0); fi;
if (k mod 2)=1 and (n mod k)=0 then return(1); fi;
if (k mod 2)=0 and ((n-k/2) mod k) = 0 then return(1); fi;
return(0);
end;
if n<(k-1)*k/2 or k>r(n) then return(0); fi;
add( (-1)^(i+1)* A237048(n, i), i=1..k); end;
MATHEMATICA
cd[n_, k_] := If[Divisible[n, k], 1, 0]; row[n_] := Floor[(Sqrt[8 n + 1] - 1)/2]; a237048[n_, k_] := If[OddQ[k], cd[n, k], cd[n - k/2, k]];
a1[n_, k_] := Sum[(-1)^(j + 1)*a237048[n, j], {j, 1, k}];
a2[n_] := Drop[FoldList[Plus, 0, Map[(-1)^(# + 1) &, Range[row[n]]] a237048[n]], 1]; Flatten[Map[a2, Range[24]]] (* data *) (* Corrected by G. C. Greubel, Apr 16 2017 *)
PROG
(PARI) t237048(n, k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0);
kmax(n) = (sqrt(1+8*n)-1)/2;
t(n, k) = sum(j=1, k, (-1)^(j+1)*t237048(n, j));
tabf(nn) = {for (n=1, nn, for (k=1, kmax(n), print1(t(n, k), ", "); ); print(); ); } \\ Michel Marcus, Sep 20 2015
CROSSREFS
Cf. A000012, A000035, A000203, A003418, A067742, A115357, A171182, A237048, A237270, A237271, A237593, A238443, A245685, A246955, A246956, A247687, A337976, A366981.
Triangle read by rows in which row n lists the widths of the symmetric representation of sigma(n).
+10
63
1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1
COMMENTS
Here T(n,k) is defined to be the "k-th width" of the symmetric representation of sigma(n), with n>=1 and 1<=k<=2n-1. Explanation: consider the diagram of the symmetric representation of sigma(n) described in A236104, A237593 and other related sequences. Imagine that the diagram for sigma(n) contains 2n-1 equidistant segments which are parallel to the main diagonal [(0,0),(n,n)] of the quadrant. The segments are located on the diagonal of the cells. The distance between two parallel segment is equal to sqrt(2)/2. T(n,k) is the length of the k-th segment divided by sqrt(2). Note that the triangle contains nonnegative terms because for some n the value of some widths is equal to zero. For an illustration of some widths see Hartmut F. W. Hoft's contribution in the Links section of A237270. Row n has length 2*n-1.
If n is a power of 2 then all terms of row n are 1's.
If n is an even perfect number then all terms of row n are 1's except the middle term which is 2.
If n is an odd prime then row n lists (n+1)/2 1's, n-2 zeros, (n+1)/2 1's.
The number of blocks of positive terms in row n gives A237271(n). The sum of the k-th block of positive terms in row n gives A237270(n,k). It appears that the trapezoidal numbers ( A165513) are also the numbers k > 1 with the property that some of the noncentral widths of the symmetric representation of sigma(k) are not equal to 1. - Omar E. Pol, Mar 04 2023
EXAMPLE
Triangle begins:
1;
1,1,1;
1,1,0,1,1;
1,1,1,1,1,1,1;
1,1,1,0,0,0,1,1,1;
1,1,1,1,1,2,1,1,1,1,1;
1,1,1,1,0,0,0,0,0,1,1,1,1;
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
1,1,1,1,1,0,0,1,1,1,0,0,1,1,1,1,1;
1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1;
1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1;
1,1,1,1,1,1,1,1,1,2,2,2,2,2,1,1,1,1,1,1,1,1,1;
...
---------------------------------------------------------------------------
. Written as an isosceles triangle Diagram of
. the sequence begins: the symmetry of sigma
---------------------------------------------------------------------------
. _ _ _ _ _ _ _ _ _ _ _ _
. 1; |_| | | | | | | | | | | |
. 1,1,1; |_ _|_| | | | | | | | | |
. 1,1,0,1,1; |_ _| _|_| | | | | | | |
. 1,1,1,1,1,1,1; |_ _ _| _|_| | | | | |
. 1,1,1,0,0,0,1,1,1; |_ _ _| _| _ _|_| | | |
. 1,1,1,1,1,2,1,1,1,1,1; |_ _ _ _| _| | _ _|_| |
. 1,1,1,1,0,0,0,0,0,1,1,1,1; |_ _ _ _| |_ _|_| _ _|
. 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1; |_ _ _ _ _| _| |
. 1,1,1,1,1,0,0,1,1,1,0,0,1,1,1,1,1; |_ _ _ _ _| | _|
. 1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1; |_ _ _ _ _ _| _ _|
. 1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1; |_ _ _ _ _ _| |
.1,1,1,1,1,1,1,1,1,2,2,2,2,2,1,1,1,1,1,1,1,1,1; |_ _ _ _ _ _ _|
...
Also consider the infinite double-staircases diagram defined in A335616. For n = 15 the diagram with first 15 levels looks like this:
.
Level "Double-staircases" diagram
. _
1 _|1|_
2 _|1 _ 1|_
3 _|1 |1| 1|_
4 _|1 _| |_ 1|_
5 _|1 |1 _ 1| 1|_
6 _|1 _| |1| |_ 1|_
7 _|1 |1 | | 1| 1|_
8 _|1 _| _| |_ |_ 1|_
9 _|1 |1 |1 _ 1| 1| 1|_
10 _|1 _| | |1| | |_ 1|_
11 _|1 |1 _| | | |_ 1| 1|_
12 _|1 _| |1 | | 1| |_ 1|_
13 _|1 |1 | _| |_ | 1| 1|_
14 _|1 _| _| |1 _ 1| |_ |_ 1|_
15 |1 |1 |1 | |1| | 1| 1| 1|
.
Starting from A196020 and after the algorithm described in A280850 and A296508 applied to the above diagram we have a new diagram as shown below: .
Level "Ziggurat" diagram
. _
6 |1|
7 _ | | _
8 _|1| _| |_ |1|_
9 _|1 | |1 1| | 1|_
10 _|1 | | | | 1|_
11 _|1 | _| |_ | 1|_
12 _|1 | |1 1| | 1|_
13 _|1 | | | | 1|_
14 _|1 | _| _ |_ | 1|_
15 |1 | |1 |1| 1| | 1|
.
The 15th row
of this seq: [1,1,1,1,1,1,1,1,0,0,0,1,1,1,2,1,1,1,0,0,0,1,1,1,1,1,1,1,1]
The 15th row
The 15th row
The 15th row
.
The number of horizontal steps (or 1's) in the successive columns of the above diagram gives the 15th row of this triangle.
For more information about the parts of the symmetric representation of sigma(n) see A237270. For more information about the subparts see A239387, A296508, A280851. More generally, it appears there is the same correspondence between the original diagram of the symmetric representation of sigma(n) and the "Ziggurat" diagram of n. (End)
MATHEMATICA
(* function segments are defined in A237270 *) a249351[n_] := Flatten[Map[segments, Range[n]]]
CROSSREFS
Cf. A000203, A003056, A067742, A071562, A165513, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A238443, A239660, A239932- A239934, A240542, A241008, A241010, A245092, A245685, A246955, A246956, A247687, A249223, A250068, A250070, A250071, A262626, A280850, A280851, A296508, A235616, A347186.
a(n) = 0 if n is of the form m*(m+3)/2, otherwise 1.
+10
59
0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1
COMMENTS
From Stark: "alpha = 0.101101110111101111101111110 ... is irrational. For if alpha were rational, its decimal expansion would be periodic and have a period of length r starting with the k-th digit of the expansion.
"But by the very nature of alpha, there will be blocks of r digits, all 1, in this expansion after the k-th digit and the periodicity would then guarantee that everything after such a block of r digits would also be all ones.
"This contradicts the fact that there will always be zeros occurring after any given point in the expansion of alpha. Hence alpha is irrational."
Sequence B is called a reverse reluctant sequence of sequence A, if B is triangle array read by rows: row number k lists first k elements of the sequence A in reverse order. A023532 is reverse reluctant sequence of sequence A211666. - Boris Putievskiy, Jan 11 2013 An example of a sequence with infinite critical exponent [Vaslet]. - N. J. A. Sloane, May 05 2013
REFERENCES
Harold M. Stark, An Introduction to Number Theory, The MIT Press, Cambridge, Mass, eighth printing 1994, page 170.
FORMULA
Blocks of lengths 1, 2, 3, 4, ... of ones separated by a single zero.
a(n) = 1 - floor((sqrt(9+8n)-1)/2) + floor((sqrt(1+8n)-1)/2). - Paul Barry, May 25 2004 a(n) = A211666(m), where m = (t^2 + 3*t + 4)/2n - n, t = floor((-1 + sqrt(8*n-7))/2). - Boris Putievskiy, Jan 11 2013
EXAMPLE
As a triangular array written by rows, the sequence begins:
0;
1, 0;
1, 1, 0;
1, 1, 1, 0;
1, 1, 1, 1, 0;
1, 1, 1, 1, 1, 0;
1, 1, 1, 1, 1, 1, 0;
...
(End)
MATHEMATICA
a = {}; Do[a = Append[a, Join[ {0}, Table[1, {n} ] ] ], {n, 1, 13} ]; a = Flatten[a]
Table[PadLeft[{0}, n, 1], {n, 0, 20}]//Flatten (* Harvey P. Dale, Jul 10 2019 *)
PROG
(Haskell)
a023532 = (1 -) . a010052 . (+ 9) . (* 8)
a023532_list = concat $ iterate (\rs -> 1 : rs) [0]
(Python)
from sympy.ntheory.primetest import is_square
Inventory sequence: record the number of zeros thus far in the sequence, then the number of ones thus far, then the number of twos thus far and so on, until a zero is recorded; the inventory then starts again, recording the number of zeros.
+10
59
0, 1, 1, 0, 2, 2, 2, 0, 3, 2, 4, 1, 1, 0, 4, 4, 4, 1, 4, 0, 5, 5, 4, 1, 6, 2, 1, 0, 6, 7, 5, 1, 6, 3, 3, 1, 0, 7, 9, 5, 3, 6, 4, 4, 2, 0, 8, 9, 6, 4, 9, 4, 5, 2, 1, 3, 0, 9, 10, 7, 5, 10, 6, 6, 3, 1, 4, 2, 0, 10, 11, 8, 6, 11, 6, 9, 3, 2, 5, 3, 2, 0, 11, 11, 10
COMMENTS
To get started we ask: how many zero terms are there? Since there are no terms in the sequence yet, we record a '0', and having recorded a '0', we begin again: How many zero terms are there? There is now one 0, so we record a '1' and continue. How many 1's are there? There's currently one '1' in the sequence, so we record a '1' and continue. How many 2's are there? There are no 2's yet, so we record a '0', and having recorded a 0, we begin again with the question "how many zero terms are there?" And so on.
a(46) = 0 because no 8's appear before it; but note a higher number, namely 9, has appeared. - Michael S. Branicky, Mar 16 2021 A similar situation occurs at n=124, where 14 has not yet appeared in the sequence, although 15 has appeared.
The first 1000 terms seem to grow more or less in saw-tooth fashion with the largest terms (= the number of 0's), as well as the distance between the 0's, both approximately equal to the inverse triangular numbers A003056 (see attached graph #1). But the picture changes when we go out to 10000 terms. Around the 1700th term, the 1's become more frequent than the 0's and the largest values are consistently somewhat larger than the inverse triangular numbers. Around the 2500th term the 2's become the most frequent number. Also after some 4000 terms, the largest values become much larger than the inverse triangular numbers. See graph #2. (End)
Comment on the colored plot of the first 1000467 terms, from Hans Havermann, May 02 2021: (Start) If one is drawing a points-joined graph, it will obscure some of the inherent large-number dynamics. To get around that, this plot joins the points with a green line, superimposing the actual points in blue. This plot was created by Mathematica.
Your browser will likely compress the very large image to window size, so click on it to expand.
The points fall into linear features of the various counts of the various integers. The count for each integer changes as we move towards infinity and hence crosses over (changes place with) other counts unpredictably.
I decided to chart (see the blue text) the twenty largest counts at the rightmost spike which runs from the zero at 997010 to the zero at 1000467. These largest values are for the counts of integers 2 to 21 and appear at a(997013) for the 2-count; a(997014) for the 3-count, ..., and a(997032) for the 21-count.
The counts are 15275, 26832, 40162, 48539, 56364, 54372, 53393, 43588, 37288, 27396, 22425, 16735, 13099, 11460, 9466, 8386, 7191, 6478, 5777, and 5208, respectively. In my text they are sorted largest-to-smallest and written "count @ integer-being-counted": 56364 @ 6, 54372 @ 7, 53393 @ 8, 48539 @ 5, ... 5208 @ 21. (End)
A useful view may be gained by plotting the sequence against itself with an offset. Using the "Plot 2" link in the web page footer, enter " A342585" as sequences 1 and 2. Select "Plot Seq2(n+shift) vs Seq1(n)" and "Draw line segments". Start with "1" as the shift. The sequence appears somewhat like a fan, the first 4 or 5 sectors showing clearly, later sectors overlying each other. Larger shift values effectively compress early sectors into the vertical axis, making later sectors more visible. - Peter Munn, May 08 2021 For a version where a row ends not at the first zero, but rather at the last zero, see A347317. - N. J. A. Sloane, Sep 10 2021 For n around 2.5*10^9, the upper envelope of the sequence seems to be growing roughly like n/50, or maybe like O(n/log(n)). - N. J. A. Sloane, Feb 10 2023
EXAMPLE
As an irregular triangle this begins:
0;
1, 1, 0;
2, 2, 2, 0;
3, 2, 4, 1, 1, 0;
4, 4, 4, 1, 4, 0;
5, 5, 4, 1, 6, 2, 1, 0;
6, 7, 5, 1, 6, 3, 3, 1, 0;
7, 9, 5, 3, 6, 4, 4, 2, 0;
8, 9, 6, 4, 9, 4, 5, 2, 1, 3, 0;
9, 10, 7, 5, 10, 6, 6, 3, 1, 4, 2, 0;
10, 11, 8, 6, 11, 6, 9, 3, 2, 5, 3, 2, 0;
...
a(1) is 0 because the count is reset, and as yet there is no zero term immediately following another term. a(2) = 1 since the count is reset, a(1) = 0 and a(0) precedes it. The count now increments to terms equal to 1.
a(3) = 1 since a(2) = 1 and a(1) precedes it. a(4) = 0 because there is no term equal to 2 which is immediately preceded by another term.
a(5) = 2 since the count is reset, a(1) = a(4) = 0 and a(0), a(3) respectively, precede them. (End)
MAPLE
a:= proc(n) option remember; local t;
t:= `if`(a(n-1)=0, 0, b(n-1)+1);
b(n):=t; add(`if`(a(j)=t, 1, 0), j=1..n-1)
end: b(1), a(1):= 0$2:
MATHEMATICA
a[n_] := a[n] = Module[{t}, t = If[a[n-1] == 0, 0, b[n-1]+1];
b[n] = t; Sum[If[a[j] == t, 1, 0], {j, 1, n-1}]];
b[1] = 0; a[1] = 0;
PROG
(Python)
def calc(required_value_number):
values_lst = []
current_count = 0
new_value = 0
for i in range(required_value_number):
new_value = values_lst.count(current_count)
values_lst.append(new_value)
if new_value == 0:
current_count = 0
else:
current_count += 1
return new_value # Written by Gilad Moyal
(Python)
from collections import Counter
def aupton(terms):
num, alst, inventory = 0, [0], Counter([0])
for n in range(2, terms+1):
c = inventory[num]
num = 0 if c == 0 else num + 1; alst.append(c); inventory.update([c])
return alst
(PARI)
A342585_vec(N, c=[], i)=vector(N, j, while(#c<=i||#c<=c[i+1], c=concat(c, 0)); c[i+=1]+if(c[1+c[i]]++&&!c[i]||j==1, i=0)) \\ M. F. Hasler, Nov 13 2021
(PARI) \\ See Links section.
(MATLAB)
function [val, arr]=invSeq(N) % val = Nth term, arr = whole array up to N
k=0;
arr=zeros(1, N); % pre-allocate array
for i=1:N
an=sum((k==arr(2:i)));
arr(i)=an;
if an == 0
k = 0;
else
k=k+1;
end
end
val=arr(end);
(R)
# Prints the first 10, 068 terms
library("dplyr")
options(max.print=11000)
inventory <- data.frame(1, 0)
colnames(inventory) <- c("n", "an")
value_to_count = 0
n = 1
for(x in 1:128) # Increase the 128 for more terms. The number of terms
# given is on the order of x^1.9 in the region around 128.
{
status <- TRUE
while(status)
{
count <- length(which(inventory$an == value_to_count))
n = n + 1
inventory <- rbind(inventory, c(n, count))
status <- isTRUE(count != 0)
value_to_count = value_to_count + 1
}
value_to_count = 0
}
The midpoint of the (rotated) Dyck path from (0, n) to (n, 0) defined by A237593 has coordinates (a(n), a(n)). Also a(n) is the alternating sum of the n-th row of A235791.
+10
54
1, 2, 2, 3, 3, 5, 5, 6, 7, 7, 7, 9, 9, 9, 11, 12, 12, 13, 13, 15, 15, 15, 15, 17, 18, 18, 18, 20, 20, 22, 22, 23, 23, 23, 25, 26, 26, 26, 26, 28, 28, 30, 30, 30, 32, 32, 32, 34, 35, 36, 36, 36, 36, 38, 38, 40, 40, 40, 40, 42, 42, 42, 44, 45, 45, 47, 47, 47, 47, 49, 49, 52, 52
COMMENTS
The sequence is closely related to the alternating harmonic series.
Its asymptotic behavior is lim_{k -> infinity} a(k)/k = log 2. The relative error is abs(a(k) - k*log(2))/(k*log(2)) <= 2/sqrt(k).
Conjecture 1: the sequence of first positions of the alternating runs of odd and even numbers in a(k) equals A028982. Example: the positions in (1),(2),2,(3),3,5,5,(6),(7),7,7,9,9,9,11,(12),12,(13),13,15,... are 1,2,4,8,9,16,18,... Checked with a Mathematica function through a(1000000). Conjecture 2: The sequence of differences a(n) - a(n-1), for n>=1, appears to be equal to A067742(n), the sequence of middle divisors of n; as an empty sum, a(0) = 0, (which was conjectured by Michel Marcus in the entry A237593). Checked with the two respective Mathematica functions up to n=100000. - Hartmut F. W. Hoft, Jul 17 2014 Conjecture 3: a(n) is also the difference between the total number of partitions of all positive integers <= n into an odd number of consecutive parts, and the total number of partitions of all positive integers <= n into an even number of consecutive parts. - Omar E. Pol, Apr 28 2017 Conjecture 4: a(n) is also the total number of central subparts of all symmetric representations of sigma of all positive integers <= n. - Omar E. Pol, Apr 29 2017 a(n) is also the sum of the odd-indexed terms of the n-th row of the triangle A237591. - Omar E. Pol, Jun 20 2018
EXAMPLE
Illustration of initial terms in two ways in accordance with the sum of the odd-indexed terms of the rows of A237591: .
n a(n) _ _
1 1 _|_| |_|_
2 2 _|_ _| |_ _|
3 2 _|_ _| |_ _|_
4 3 _|_ _ _| |_ _ _|
5 3 _|_ _ _| _ |_ _ _|_ _
6 5 _|_ _ _ _| |_| |_ _ _ _|_|
7 5 _|_ _ _ _| |_| |_ _ _ _|_|_
8 6 _|_ _ _ _ _| _|_| |_ _ _ _ _|_|_
9 7 _|_ _ _ _ _| |_ _| |_ _ _ _ _|_ _|
10 7 _|_ _ _ _ _ _| |_| |_ _ _ _ _ _|_|
11 7 _|_ _ _ _ _ _| _|_| |_ _ _ _ _ _|_|_ _
12 9 _|_ _ _ _ _ _ _| |_ _| |_ _ _ _ _ _ _|_ _|
13 9 _|_ _ _ _ _ _ _| |_ _| |_ _ _ _ _ _ _|_ _|
14 9 _|_ _ _ _ _ _ _ _| _|_| _ |_ _ _ _ _ _ _ _|_|_ _
15 11 _|_ _ _ _ _ _ _ _| |_ _| |_| |_ _ _ _ _ _ _ _|_ _|_|_
16 12 |_ _ _ _ _ _ _ _ _| |_ _| |_| |_ _ _ _ _ _ _ _ _|_ _|_|
...
Figure 1. Figure 2.
.
Figure 1 shows the illustration of initial terms taken from the isosceles triangle of A237593. For n = 16 there are (9 + 2 + 1) = 12 cells in the 16th row of the diagram, so a(16) = 12. Figure 2 shows the illustration of initial terms taken from an octant of the pyramid described in A244050 and A245092. For n = 16 there are (9 + 2 + 1) = 12 cells in the 16th row of the diagram, so a(16) = 12. Note that if we fold each level (or row) of that isosceles triangle of A237593 we essentially obtain the structure of the pyramid described in A245092 whose terraces at the n-th level have a total area equal to sigma(n) = A000203(n). (End).
MATHEMATICA
a[n_] := Sum[(-1)^(k + 1) Ceiling[(n + 1)/k - (k + 1)/2], {k, 1, Floor[-1/2 + 1/2 Sqrt[8 n + 1]]}]; Table[a[n], {n, 40}]
PROG
(PARI) a(n) = sum(k=1, floor(-1/2 + 1/2*sqrt(8*n + 1)), (-1)^(k + 1)*ceil((n + 1)/k - (k + 1)/2)); \\ Indranil Ghosh, Apr 21 2017 (Python)
from sympy import sqrt
import math
def a(n): return sum((-1)**(k + 1) * int(math.ceil((n + 1)/k - (k + 1)/2)) for k in range(1, int(math.floor(-1/2 + 1/2*sqrt(8*n + 1))) + 1))
CROSSREFS
Cf. A028982, A067742, A196020, A236104, A235791, A237270, A237271, A237591, A237593, A071562, A259176, A259179, A279387, A286001, A322141, A338204, A338758.
Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with k modes.
+10
54
1, 0, 1, 0, 2, 0, 2, 1, 0, 4, 1, 0, 5, 2, 0, 7, 3, 1, 0, 11, 3, 1, 0, 16, 4, 2, 0, 21, 6, 3, 0, 29, 8, 4, 1, 0, 43, 7, 5, 1, 0, 54, 13, 8, 2, 0, 78, 12, 8, 3, 0, 102, 17, 11, 5, 0, 131, 26, 12, 6, 1, 0, 175, 29, 17, 9, 1, 0, 233, 33, 18, 11, 2, 0, 295, 47, 25
COMMENTS
A mode in a multiset is an element that appears at least as many times as each of the others. For example, the modes of {a,a,b,b,b,c,d,d,d} are {b,d}.
EXAMPLE
Triangle begins:
1
0 1
0 2
0 2 1
0 4 1
0 5 2
0 7 3 1
0 11 3 1
0 16 4 2
0 21 6 3
0 29 8 4 1
0 43 7 5 1
0 54 13 8 2
0 78 12 8 3
0 102 17 11 5
0 131 26 12 6 1
0 175 29 17 9 1
Row n = 8 counts the following partitions:
(8) (53) (431)
(44) (62) (521)
(332) (71)
(422) (3311)
(611)
(2222)
(3221)
(4211)
(5111)
(22211)
(32111)
(41111)
(221111)
(311111)
(2111111)
(11111111)
MATHEMATICA
msi[ms_]:=Select[Union[ms], Count[ms, #]>=Max@@Length/@Split[ms]&];
Table[Length[Select[IntegerPartitions[n], Length[msi[#]]==k&]], {n, 0, 15}, {k, 0, Floor[(Sqrt[1+8n]-1)/2]}]
CROSSREFS
Removing columns 0 and 1 and taking sums gives A362607, ranks A362605. This statistic (mode-count) is ranked by A362611. A008284 counts partitions by length.
A275870 counts collapsible partitions.
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