A275444 Triangle read by rows: T(n,k) is the number of compositions of n with parts in {1,2,3} and having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/3)).
1, 1, 2, 2, 2, 3, 4, 3, 10, 6, 14, 4, 6, 26, 12, 11, 34, 36, 11, 62, 68, 8, 20, 82, 140, 32, 20, 144, 228, 112, 37, 186, 424, 264, 16, 37, 316, 664, 608, 80, 68, 404, 1176, 1168, 320, 68, 676, 1784, 2312, 896, 32, 125, 860, 3032, 4096, 2304, 192
Offset: 0
Examples
Row 4 is [3,4] because the compositions of 4 with parts in {1,2,3} are 13, 31, 22, 211, 121, 112, and 1111, having asymmetry degrees 1, 1, 0, 1, 0, 1, and 0, respectively.
Triangle starts:
1;
1;
2;
2,2;
3,4;
3,10;
6,14,4.
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)*(1+z^2)/(1-z^2-2*t*z^3-(1+2*t)*z^4-2*t*z^5-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, Ceiling[Length[#]/2]], Reverse@ Take[# - 1, -Ceiling[Length[#]/2]]]] &, Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {a_, _} /; a > 3]], 1]]], {n, 0, 17}] // Flatten (* Michael De Vlieger, Aug 17 2016 *)
Formula
G.f.: G(t,z) = (1+z)*(1+z^2)/(1-z^2-z^4-z^6-2*t*z^3*(1+z+z^2)). 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