A275446 Triangle read by rows: T(n,k) is the number of compositions of n with parts in {2,1,3,5,7,9,...} and having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/3)).
1, 1, 2, 2, 2, 3, 4, 4, 10, 6, 16, 4, 8, 30, 12, 11, 48, 36, 15, 82, 76, 8, 21, 128, 164, 32, 29, 204, 312, 112, 40, 312, 596, 288, 16, 55, 482, 1064, 704, 80, 76, 728, 1884, 1536, 320, 105, 1100, 3212, 3248, 960, 32, 145, 1640, 5428, 6464, 2624, 192
Offset: 0
Examples
Row 4 is [3,4] because the compositions of 4 with parts in {2,1,3,5,7,...} are 22, 31, 13, 211, 121, 112, and 1111, having asymmetry degrees 0, 1, 1, 1, 0, 1, and 0, respectively. Triangle starts: 1; 1; 2; 2,2; 3,4; 4,10; 6,16,4.
References
- S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
Links
- Krithnaswami Alladi and V. E. Hoggatt, Jr. Compositions with Ones and Twos, Fibonacci Quarterly, 13 (1975), 233-239.
- 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^4)/(1-z-z^2+(1-2*t)*z^3-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/3], 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_, _} /; And[EvenQ@ a, a != 2]]], 1]]], {n, 0, 16}] // Flatten (* Michael De Vlieger, Aug 17 2016 *)
Formula
G.f.: G(t,z) = (1-z^4)/(1-z-z^2+(1-2*t)*z^3-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