A297120 Number of compositions derived from the overpartitions of n.

1, 2, 5, 14, 36, 92, 234, 586, 1452, 3562, 8674, 20956, 50290, 119922, 284308, 670458, 1573250, 3674700, 8546282, 19796234, 45681908, 105041402, 240723618, 549919604, 1252492674, 2844551866, 6442833156, 14555300218, 32801922154, 73749649900, 165443000338

Offset: 0

Author

Gregory L. Simay, Dec 25 2017

Keywords

• nonn

Comments

Start by enumerating the overpartitions of n, then allow the parts to vary their arrangements.

Examples

			The A015128(4) = 14 overpartitions of 4 are: 4; 4'; 3,1; 3,1'; 3'1; 3',1', 2,2; 2',2; 2,1,1; 2,1',1; 2',1,1; 2',1',1; 1,1,1,1; and 1',1,1,1.  The corresponding 36 compositions are 4; 4'; 3,1; 1,3; 3,1'; 1',3; 3',1; 1,3'; 3',1'; 1',3'; 2,2; 2,2'; 2',2; 2,1,1; 1,2,1; 1,1,2; 2,1,1'; 2,1',1; 1,2,1'; 1,1',2'; 1',1,2; 1',2,1; 2',1,1; 1,2',2; 1,1,2'; 2',1,1'; 2',1',1; 1,2',1'; 1,1',2'; 1',2',1; 1',1,2'; 1,1,1,1; 1,1,1,1'; 1,1,1',1; 1,1',1,1; and 1',1,1,1. Note: For a sequence of like parts p,p,...p, an overcomposition of n will only recognize p,p...p and p',p...,p; the p' is not allowed to be other than the initial p term.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..3000 (terms 0..1000 from Alois P. Heinz)

Crossrefs

Cf. A015128, A236002.

Programs

Maple

b:= proc(n, i, p) option remember; `if`(n=0 or i=1, (p+n)!*
      (1+n)/n!, add(b(n-i*j, i-1, p+j)*(1+j)/j!, j=0..n/i))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..35);  # Alois P. Heinz, Dec 26 2017

Mathematica

b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1, (p + n)!*(n + 1)/n!, Sum[b[n - i*j, i - 1, p + j]*(j + 1)/j!, {j, 0, n/i}]];
a[n_] := b[n, n, 0];
Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Dec 27 2017, after Alois P. Heinz *)

PARI

{my(n=30); apply(p->subst(serlaplace(p), y, 1), Vec(prod(k=1, n, (1+y*x^k)*exp(y*x^k + O(x*x^n)))))} \\ Andrew Howroyd, Dec 26 2017

Python

from sympy.core.cache import cacheit
from sympy import factorial
@cacheit
def b(n, i, p):  return factorial(p + n)*(n + 1)//factorial(n) if n==0 or i==1 else sum(b(n - i*j, i - 1, p + j)*(j + 1)//factorial(j) for j in range(n//i + 1))
def a(n): return b(n, n, 0)
print([a(n) for n in range(41)]) # Indranil Ghosh, Dec 29 2017, after Maple code

Extensions

More terms from Alois P. Heinz, Dec 26 2017