A319398 Number of partitions of n into exactly five positive Fibonacci numbers.
0, 0, 0, 0, 0, 1, 1, 2, 2, 4, 4, 5, 5, 7, 6, 7, 7, 8, 7, 9, 8, 10, 9, 9, 8, 11, 8, 10, 10, 11, 10, 11, 10, 13, 10, 11, 8, 10, 10, 10, 11, 12, 11, 11, 11, 13, 11, 12, 11, 12, 12, 11, 11, 13, 12, 10, 8, 10, 9, 9, 12, 11, 10, 13, 10, 14, 14, 11, 11, 11, 11, 13
Offset: 0
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..17711
Programs
-
Maple
h:= proc(n) option remember; `if`(n<1, 0, `if`((t-> issqr(t+4) or issqr(t-4))(5*n^2), n, h(n-1))) end: b:= proc(n, i, t) option remember; `if`(n=0, 1, `if`(i<1 or t<1, 0, b(n, h(i-1), t)+b(n-i, h(min(n-i, i)), t-1))) end: a:= n-> (k-> b(n, h(n), k)-b(n, h(n), k-1))(5): seq(a(n), n=0..120);
-
Mathematica
h[n_] := h[n] = If[n < 1, 0, If[Function[t, IntegerQ@Sqrt[t + 4] || IntegerQ@Sqrt[t - 4]][5 n^2], n, h[n - 1]]]; b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1, If[i < 1 || t < 1, 0, b[n, h[i - 1], t] + b[n - i, h[Min[n - i, i]], t - 1]]]; a[n_] := With[{k = 5}, b[n, h[n], k] - b[n, h[n], k - 1]]; a /@ Range[0, 120] (* Jean-François Alcover, Dec 08 2020 *)
Formula
a(n) = [x^n y^5] 1/Product_{j>=2} (1-y*x^A000045(j)).