A258458 Number of partitions of n into parts of exactly 3 sorts which are introduced in ascending order.
1, 7, 33, 130, 463, 1557, 5031, 15877, 49240, 151116, 460173, 1394645, 4212071, 12693724, 38195286, 114817389, 344911117, 1035659955, 3108817911, 9330152740, 27997803871, 84008165515, 252053831034, 756220333901, 2268778132337, 6806569134920, 20420175154486
Offset: 3
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 3..1000
Programs
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Maple
b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0, b(n, i-1, k) +`if`(i>n, 0, k*b(n-i, i, k)))) end: T:= (n, k)-> add(b(n$2, k-i)*(-1)^i/(i!*(k-i)!), i=0..k): a:= n-> T(n,3): seq(a(n), n=3..35);
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Mathematica
b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i < 1, 0, b[n, i - 1, k] + If[i > n, 0, k*b[n - i, i, k]]]]; T[n_, k_] := Sum[b[n, n, k - i]*(-1)^i/(i!*(k - i)!), {i, 0, k}]; a[n_] := T[n, 3]; Table[a[n], {n, 3, 35}] (* Jean-François Alcover, May 22 2018, translated from Maple *)
Formula
a(n) ~ c * 3^n, where c = 1/(6*Product_{n>=1} (1-1/3^n)) = 1/(6*QPochhammer[1/3, 1/3]) = 1/(6*A100220) = 0.297552056999755698394581... . - Vaclav Kotesovec, Jun 01 2015