A276060 Triangle read by rows: T(n,k) is the number of compositions of n into parts congruent to 1 mod 3 and having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/5)).
1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 4, 3, 6, 3, 10, 5, 10, 4, 4, 20, 4, 7, 22, 12, 6, 34, 20, 10, 42, 36, 9, 64, 48, 8, 15, 70, 96, 8, 13, 112, 120, 32, 22, 124, 204, 56, 19, 184, 280, 112, 32, 212, 436, 176, 16, 28, 310, 564, 360, 16, 47, 346, 896, 504, 80, 41, 512, 1128, 920, 144, 69, 570, 1704, 1360
Offset: 0
Examples
Row 7 is [2,4] because the compositions of 7 with parts in {1,4,7,10,...} are 7, 4111, 1411, 1141, 1114, and 1111111, having asymmetry degrees 0, 1, 1, 1, 1, and 0, respectively. Triangle starts: 1; 1; 1; 1; 2; 1,2; 2,2.
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^2)*(1+z+z^2)*(1-z+z^2)*(1+z-z^3)/(1-z^2-z^3+z^5-2*t*z^5-z^6+z^9): 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[TakeWhile[BinCounts[#, {0, 1 + Floor[n/5], 1}], # != 0 &] &@ 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_, _} /; Mod[a, 3] != 1]], 1]]], {n, 0, 24}] // Flatten (* Michael De Vlieger, Aug 22 2016 *)
Formula
G.f.: G(t,z) = (1-z^2)*(1+z+z^2)*(1-z+z^2)*(1+z-z^3)/(1-z^2-z^3+z^5-2*t*z^5-z^6+z^9). 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