Displaying 11-15 of 15 results found.
Triangle read by rows: T(n,k) is the number of compositions of n with parts in {1,2} having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/3)).
+10
1
1, 1, 2, 1, 2, 3, 2, 2, 6, 5, 4, 4, 3, 14, 4, 8, 10, 16, 5, 30, 12, 8, 13, 20, 48, 8, 8, 60, 36, 40, 21, 40, 124, 32, 16, 13, 116, 88, 144, 16, 34, 76, 292, 112, 96, 21, 218, 204, 432, 80, 32, 55, 142, 648, 320, 400, 32, 34, 402, 444, 1160, 320, 224
COMMENTS
The asymmetry degree of a finite sequence of numbers is defined to be the number of pairs of symmetrically positioned distinct entries. Example: the asymmetry degree of (2,7,6,4,5,7,3) is 2, counting the pairs (2,3) and (6,5).
Number of entries in row n is 1 + floor(n/3).
Sum of entries in row n is A000045(n+1) (Fibonacci). T(n,0) = A053602(n+1) (= number of palindromic compositions of n with parts in {1,2}).
FORMULA
G.f.: G(t,z) = (1+z+z^2)/(1-z^2-2tz^3-z^4). In the more general situation of compositions into a[1]<a[2]<a[3]<..., denoting F(z) = Sum(z^{a[j]},j>=1}, we have G(t,z)=(1+F(z))/(1-F(z^2)-t(F(z)^2-F(z^2))). In particular, for t=0 we obtain Theorem 1.2 of the Hoggatt et al. reference.
EXAMPLE
Row 4 is [3,2] because the compositions of 4 with parts in {1,2} are 22, 112, 121, 211, and 1111, having asymmetry degrees 0, 1, 0, 1, 0, respectively.
Triangle starts:
1;
1;
2;
1,2;
3,2;
2,6;
5,4,4.
MAPLE
G:=(1+z+z^2)/(1-z^2-2*t*z^3-z^4): Gser:=simplify(series(G, z=0, 25)): for n from 0 to 20 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 20 do seq(coeff(P[n], t, j), j=0..degree(P[n])) end do; # yields sequence in triangular form
MATHEMATICA
Join[{{1}}, Table[BinCounts[#, {0, 1 + Floor[n/3], 1}] &@ Map[Total, Map[BitXor[Take[# - 1, Ceiling[Length[#]/2]], Reverse@ Take[# - 1, -Ceiling[Length[#]/2]]] &, Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {a_, ___} /; a > 2]], 1]]], {n, 17}]] // Flatten (* Michael De Vlieger, Aug 17 2016 *)
Triangle read by rows: T(n,k) is the number of compositions of n into odd parts, having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/4)).
+10
1
1, 1, 1, 2, 1, 2, 3, 2, 2, 6, 5, 8, 3, 14, 4, 8, 22, 4, 5, 30, 20, 13, 52, 24, 8, 60, 68, 8, 21, 112, 92, 8, 13, 116, 192, 56, 34, 228, 284, 64, 21, 218, 484, 248, 16, 55, 446, 768, 312, 16, 34, 402, 1132, 872, 144, 89, 848, 1900, 1184, 160, 55, 730
COMMENTS
The asymmetry degree of a finite sequence of numbers is defined to be the number of pairs of symmetrically positioned distinct entries. Example: the asymmetry degree of (2,7,6,4,5,7,3) is 2, counting the pairs (2,3) and (6,5).
Number of entries in row n is 1 + floor(n/4).
Sum of entries in row n is A000045(n) (Fibonacci). T(n,0) = A053602(n) (= number of palindromic compositions of n into odd parts).
REFERENCES
S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
FORMULA
G.f.: G(t,z) = (1-z^4)(1+z-z^2)/(1-2z^2-2tz^4+z^6). In the more general situation of compositions into a[1]<a[2]<a[3]<..., denoting F(z) = Sum(z^{a[j]},j>=1}, we have G(t,z) =(1 + F(z))/(1 - F(z^2) - t(F(z)^2 - F(z^2))). In particular, for t=0 we obtain Theorem 1.2 of the Hoggatt et al. reference.
EXAMPLE
Row 5 is [3,2] because the compositions of 5 into odd parts are 5, 113, 131, 311, and 11111, having asymmetry degrees 0, 1, 0, 1, 0, respectively.
Triangle starts:
1;
1;
1;
2;
1,2;
3,2;
2,6;
5,8; .
MAPLE
G := (1-z^4)*(1+z-z^2)/(1-2*z^2-2*t*z^4+z^6): Gser := simplify(series(G, z = 0, 30)): for n from 0 to 25 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 25 do seq(coeff(P[n], t, j), j = 0 .. degree(P[n])) end do; # yields sequence in triangular form
MATHEMATICA
Table[BinCounts[#, {0, 1 + Floor[n/4], 1}] &@ Map[Total, Map[Map[Boole[# >= 1] &, BitXor[Take[(# - 1)/2, Ceiling[Length[#]/2]], Reverse@ Take[(# - 1)/2, -Ceiling[Length[#]/2]]]] &, Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {___, a_, ___} /; EvenQ@ a]], 1]]], {n, 0, 20}] // Flatten (* Michael De Vlieger, Aug 17 2016 *)
0, 0, 1, 2, 2, 3, 4, 6, 7, 10, 12, 17, 20, 28, 33, 46, 54, 75, 88, 122, 143, 198, 232, 321, 376, 520, 609, 842, 986, 1363, 1596, 2206, 2583, 3570, 4180, 5777, 6764, 9348, 10945, 15126, 17710, 24475, 28656, 39602, 46367, 64078, 75024, 103681, 121392, 167760, 196417, 271442
COMMENTS
Sequences A000071 and A001610 look like long-lost cousins, and this entry smoothly interleaves them. Differences between successive terms are Fibonacci numbers.
FORMULA
a(n) = Fibonacci(k+2) + Fibonacci(k)*(n mod 2) - 1, where k = floor(n/2). - Wesley Ivan Hurt, Dec 09 2020
MATHEMATICA
Block[{b = {0, 2}, a = {}}, Do[If[EvenQ[i], AppendTo[b, Total@ b[[-2 ;; -1]] + 1 ]; AppendTo[a, Fibonacci[i/2 + 1] - 1], AppendTo[a, b[[(i - 1)/2]]]], {i, 2, 53}]; a] (* Michael De Vlieger, Dec 09 2020 *)
CROSSREFS
The first differences are essentially A053602.
Find the largest k with F(k) <= n, where F(k) is the k-th Fibonacci number. a(n) = F(k+2) + n.
+10
1
1, 4, 7, 11, 12, 18, 19, 20, 29, 30, 31, 32, 33, 47, 48, 49, 50, 51, 52, 53, 54, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 199, 200, 201, 202, 203, 204
COMMENTS
The terms consist of 1 together with numbers that appear in row m of the Wythoff array ( A035513) if m is in the sequence. a(0) = 1, otherwise a(n) is the number whose Zeckendorf representation is "10" followed by the Zeckendorf representation of n.
If we define an extended Zeckendorf representation to be the Zeckendorf representation with "01" appended, then the numbers in the sequence are exactly those whose extended representation starts 101... . This extended representation is a valid Fibonacci base representation if we specify the rightmost digit to have weight F(0) = 0.
Equivalently, for positive integer m, find the largest k with F(k) <= m, where F(k) is the k-th Fibonacci number. m is in the sequence if and only if m >= F(k) + F(k-2).
Numbers given to rabbits on Rabbit 1's branch of the generation tree described in the A035513 examples. Equivalently, take the positive integers in turn, placing runs of them alternatively into 2 sets, with run lengths from A053602/ A051792 (self-interleaved Fibonacci sequence) as follows: set A: 1 0 1 1 2 3 5 ...
set B: 1 1 2 3 5 8 ...
The sequence lists the numbers in set A.
FORMULA
Union_{k >= 2} {m : F(k)+F(k-2) <= m < F(k+1)}, where F(k) = A000045(k).
EXAMPLE
The initial Fibonacci numbers are F(0)..F(5) = 0, 1, 1, 2, 3, 5.
For n = 0, the largest k with F(k) <= 0 is k = 0, so F(k+2) = F(2) = 1, so a(0) = 1 + 0 = 1.
For n = 1, the largest k with F(k) <= 1 is k = 2, so F(k+2) = F(4) = 3, so a(1) = 3 + 1 = 4.
For n = 4, the largest k with F(k) <= 4 is k = 4, so F(k+2) = F(6) = 8, so a(4) = 8 + 4 = 12.
In the paragraph that follows we use the Wythoff array-based definition from the start of the comments.
Every positive integer appears once (only) in the Wythoff array. 0 is not positive, so does not appear in the array, so is not in the sequence. 1 is in the sequence by definition. 2 appears in Wythoff row 0, and 0 is not in the sequence, so 2 is not in the sequence. 4 appears in Wythoff row 1, and 1 is in the sequence, so 4 is in the sequence.
MATHEMATICA
kmax=12; Flatten[Table[Range[Fibonacci[k]+Fibonacci[k-2], Fibonacci[k+1]-1], {k, 2, kmax}]] (* Paolo Xausa, Jan 02 2022 *) A108852[n_]:=1+Floor[Log[GoldenRatio, 1+n*Sqrt[5]]];
nterms=100; Table[n+Fibonacci[1+ A108852[n]], {n, 0, nterms-1}](* Paolo Xausa, Jan 02 2022 *)
PROG
(PARI) a(n) = my(k=0); while(fibonacci(k)<=n, k=k+1); n+fibonacci(k+1)
Fibonacci-Chebyshev sequence depending on rabbit sequence A005614.
+10
0
0, 1, 1, 2, 1, 3, 2, 5, 7, 2, 9, 11, 2, 13, 11, 24, 35, 11, 46, 35, 81, 116, 35, 151, 186, 35, 221, 186, 407, 593, 186, 779, 965, 186, 1151, 965, 2116, 3081, 965, 4046, 3081, 7127, 10208, 3081, 13289, 16370, 3081, 19451, 16370, 35821, 52191, 16370, 68561, 52191, 120752
COMMENTS
See Viswanath sequences ( A078416 and links) for random sequences of this type.
FORMULA
a(n) = a(n-1) + (-1)^(h(n)+1)*a(n-2), with h(n):= A005614(n) (rabbit sequence or Fibonacci word), inputs a(0)=0, a(1)=1.
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