A119907 Number of partitions of n such that if k is the largest part, then k-2 occurs as a part.

0, 0, 0, 0, 1, 1, 3, 4, 7, 9, 15, 18, 27, 34, 47, 58, 79, 96, 127, 155, 199, 242, 308, 371, 465, 561, 694, 833, 1024, 1223, 1491, 1778, 2150, 2556, 3076, 3642, 4359, 5151, 6133, 7225, 8570, 10066, 11892, 13937, 16401, 19173, 22495, 26228, 30676, 35692, 41620

Offset: 0

Author

Vladeta Jovovic, Aug 02 2006

Keywords

• easy
• nonn

Comments

It appears that positive terms give column 3 of triangle A210945. - Omar E. Pol, May 18 2012

Links

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

Crossrefs

Cf. A083751.

Programs

Maple

b:= proc(n, i) option remember;
      `if`(n=0 or i=1, 1, b(n, i-1)+`if`(i>n, 0, b(n-i, i)))
    end:
a:= n-> add(b(n-(2*k-2), k), k=3..1+n/2):
seq(a(n), n=0..60);  # Alois P. Heinz, May 18 2012

Mathematica

b[n_, i_] := b[n, i] = If[n==0 || i==1, 1, b[n, i-1] + If[i>n, 0, b[n-i, i]]]; a[n_] := Sum[b[n-(2*k-2), k], {k, 3, 1+n/2}]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Jul 01 2015, after Alois P. Heinz *)

Formula

G.f. for number of partitions of n such that if k is the largest part, then k-m occurs as a part is Sum(x^(2*i-m)/Product(1-x^j, j=1..i), i=m+1..infinity).

It appears that a(n) = (A000041(n+2) - A000041(n+1)) - (A002620(n+2) - A002620(n+1)). - Gionata Neri, Apr 12 2015

Extensions

More terms from Joshua Zucker, Aug 14 2006