A275442 Triangle read by rows: T(n,k) is the number of compositions without 2's and having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/4)).
1, 1, 1, 2, 2, 2, 3, 4, 4, 8, 5, 16, 7, 26, 4, 9, 44, 12, 12, 70, 32, 16, 108, 76, 21, 166, 156, 8, 28, 248, 308, 32, 37, 368, 572, 104, 49, 540, 1020, 288, 65, 784, 1768, 696, 16, 86, 1132, 2976, 1568, 80, 114, 1622, 4908, 3304, 304, 151, 2312, 7944, 6624, 960, 200, 3280, 12652, 12768, 2640, 32
Offset: 0
Examples
Row 5 is [3,4] because the compositions of 5 without 2's are 5, 113, 131, 311, 14, 41, and 11111, having asymmetry degrees 0, 1, 0, 1, 1, 1, and 0, respectively. Triangle starts: 1; 1; 1; 2; 2,2; 3,4; 4,8; 5,16.
References
- S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
Links
- John Tyler Rascoe, Rows n = 0..200, flattened
- Krithnaswami Alladi and V. E. Hoggatt, Jr. Compositions with Ones and Twos, Fibonacci Quarterly, 13 (1975), 233-239.
- P. Chinn and S. Heubach, Integer Sequences Related to Compositions without 2's, J. Integer Seqs., Vol. 6, 2003.
- V. E. Hoggatt, Jr. and Marjorie Bicknell, Palindromic compositions, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.
Programs
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Maple
G := (1-z^2)/(1-z-z^2+z^4-2*t*z^4): 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
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Mathematica
Table[BinCounts[#, {0, 1 + Floor[n/4], 1}] &@ Map[Total, Map[Map[Boole[# >= 1] &, BitXor[Take[# - 1, Ceiling[Length[#]/2]], Reverse@ Take[# - 1, -Ceiling[Length[#]/2]]]] &, Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {_, a_, _} /; a == 2]], 1]]], {n, 0, 20}] // Flatten (* Michael De Vlieger, Aug 17 2016 *) -
PARI
T_zt(max_row) = {my(N = max_row+1, z='z+O('z^N), h=(1-z^2)/(1-z-z^2+z^4-2*t*z^4)); vector(N, n, Vecrev(polcoeff(h, n-1)))} T_zt(10) \\ John Tyler Rascoe, May 09 2025
Formula
G.f.: G(t,z) = (1-z^2)/(1-z-z^2+z^4-2*t*z^4). In the more general situation of compositions into a[1]=1} z^(a[j]), 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.
Comments