A276064 Triangle read by rows: T(n,k) is the number of compositions of n with parts in {1,5} and having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/6)).
1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 4, 2, 4, 2, 6, 3, 8, 3, 8, 4, 4, 12, 4, 4, 10, 12, 6, 16, 12, 5, 16, 24, 8, 24, 28, 6, 26, 40, 8, 10, 36, 52, 8, 8, 40, 60, 32, 13, 56, 84, 32, 11, 58, 96, 80, 17, 84, 136, 88, 15, 80, 160, 160, 16, 23, 120, 220, 192, 16
Offset: 0
Examples
Row 8 is [1,4] because the compositions of 8 with parts in {1,5} are 5111, 1511, 1151, 1115 and 11111111, having asymmetry degrees 1, 1, 1, 1, and 0, respectively.
Triangle starts:
1;
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
-
Maple
G := (1+z+z^5)/(1-z^2-2*t*z^6-z^10): 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
-
Mathematica
Table[TakeWhile[BinCounts[#, {0, 1 + Floor[n/4], 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_, _} /; Nor[a == 1, a == 5]]], 1]]], {n, 0, 25}] // Flatten (* Michael De Vlieger, Aug 22 2016 *)
Formula
G.f.: G(t,z) = (1+z+z^5)/(1-z^2-2*t*z^6-z^10). 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