A319799 Number of partitions of 2n into exactly n positive triangular numbers.
1, 0, 1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 3, 4, 3, 5, 5, 7, 5, 7, 7, 9, 9, 9, 11, 12, 14, 14, 14, 17, 17, 21, 20, 23, 24, 27, 28, 31, 32, 36, 37, 42, 43, 47, 50, 53, 58, 61, 64, 69, 72, 82, 83, 91, 92, 102, 107, 115, 118, 128, 135, 147, 152, 159, 169, 181
Offset: 0
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..3000
Programs
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Maple
h:= proc(n) option remember; `if`(n<1, 0, `if`(issqr(8*n+1), n, h(n-1))) end: b:= proc(n, i, k) option remember; `if`(n=0, `if`(k=0, 1, 0), `if`(i*k
n, 0, b(n, h(i-1), k)+b(n-i, h(min(n-i, i)), k-1))) end: a:= n-> b(2*n, h(2*n), n): seq(a(n), n=0..80); -
Mathematica
h[n_] := h[n] = If[n < 1, 0, If[IntegerQ@Sqrt[8*n + 1], n, h[n - 1]]]; b[n_, i_, k_] := b[n, i, k] = If[n == 0, If[k == 0, 1, 0], If[i*k < n || k > n, 0, b[n, h[i - 1], k] + b[n - i, h[Min[n - i, i]], k - 1]]]; a[n_] := b[2n, h[2n], n]; a /@ Range[0, 80] (* Jean-François Alcover, Mar 12 2021, after Alois P. Heinz *)
Formula
a(n) = [x^(2n) y^n] 1/Product_{j>=1} (1-y*x^A000217(j)).
a(n) = A319797(2n,n).
G.f.: Product_{k>=1} 1 / (1 - x^(k*(k + 3)/2)). - Ilya Gutkovskiy, Jan 31 2021