A226875 Number of n-length words w over a 5-ary alphabet {a1,a2,...,a5} such that #(w,a1) >= #(w,a2) >= ... >= #(w,a5) >= 0, where #(w,x) counts the letters x in word w.

1, 1, 3, 10, 47, 246, 882, 3921, 18223, 84790, 432518, 1863951, 8892842, 42656147, 204204353, 1025014815, 4728033983, 22948258742, 111605089014, 541696830843, 2708218059022, 12861557284425, 62938669549583, 308273057334413, 1508708926286914, 7533652902408071

Offset: 0

Author

Alois P. Heinz, Jun 21 2013

Keywords

• nonn

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Column k=5 of A226873.

Cf. A027306, A092255, A092429.

Programs

Maple

b:= proc(n, i, t) option remember;
      `if`(t=1, 1/n!, add(b(n-j, j, t-1)/j!, j=i..n/t))
    end:
a:= n-> n!*b(n, 0, 5):
seq(a(n), n=0..30);

Mathematica

Table[Sum[Sum[Sum[Sum[Sum[If[i+j+k+l+m==n,n!/i!/j!/k!/l!/m!,0],{m,0,l}],{l,0,k}],{k,0,j}],{j,0,i}],{i,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jul 01 2013 *)
CoefficientList[Series[(HypergeometricPFQ[{},{},x]^5 + 10*HypergeometricPFQ[{},{},x]^3*HypergeometricPFQ[{},{1},x^2] + 20*HypergeometricPFQ[{},{},x]^2*HypergeometricPFQ[{},{1,1},x^3] + 20*HypergeometricPFQ[{},{1},x^2]*HypergeometricPFQ[{},{1,1},x^3] + 15*HypergeometricPFQ[{},{1},x^2]^2*HypergeometricPFQ[{},{},x] + 30*HypergeometricPFQ[{},{1,1,1},x^4]*HypergeometricPFQ[{},{},x] + 24*HypergeometricPFQ[{},{1,1,1,1},x^5])/5!,{x,0,20}],x]*Range[0,20]! (* more efficient, Vaclav Kotesovec, Jul 01 2013 *)

Formula

Conjecture: a(n) ~ 5^n/5!. - Vaclav Kotesovec, Mar 07 2014