A275440 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)).
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
Offset: 0
Examples
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.
References
- S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
Links
- K. Alladi and V. E. Hoggatt, Jr. Compositions with Ones and Twos, Fibonacci Quarterly, 13 (1975), 233-239.
- V. E. Hoggatt, Jr. and M. Bicknell, Palindromic compositions, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.
Programs
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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
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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 *)
Formula
G.f.: G(t,z) = (1-z^4)*(1+z-z^2)/(1-2*z^2-2*t*z^4+z^6). 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