A239242 Number of partitions of n into distinct parts for which (number of odd parts) > (number of even parts).

0, 1, 0, 1, 1, 1, 2, 1, 4, 2, 6, 3, 9, 5, 12, 9, 17, 14, 22, 22, 29, 33, 38, 48, 50, 68, 65, 95, 86, 128, 113, 172, 149, 226, 197, 295, 260, 379, 342, 485, 449, 613, 587, 773, 762, 967, 987, 1206, 1269, 1497, 1623, 1855, 2063, 2289, 2610, 2823, 3280, 3471

Offset: 0

Author

Clark Kimberling, Mar 13 2014

Keywords

• nonn
• easy

Comments

a(n) = Sum_{k>=1} A240021(n,k). - Alois P. Heinz, Apr 02 2014

Examples

			a(8) = 4 counts these partitions:  71, 53, 521, 431.
		

Links

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

Crossrefs

Cf. A239239, A239240, A239241, A239243, A000009.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n>i*(i+1)/2, 0,
     `if`(n=0, `if`(t>0, 1, 0 ), b(n, i-1, t)+`if`(i>n, 0,
      b(n-i, i-1, t+`if`(irem(i, 2)=1, 1, -1)))))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..60);  # Alois P. Heinz, Mar 15 2014

Mathematica

z = 55; p[n_] := p[n] = IntegerPartitions[n]; d[u_] := d[u] = DeleteDuplicates[u]; g[u_] := g[u] = Length[u];
Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] < Count[#, ?EvenQ] &]], {n, 0, z}] (* A239239 *)
Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] <= Count[#, ?EvenQ] &]], {n, 0, z}] (* A239240 *)
Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] == Count[#, ?EvenQ] &]], {n, 0, z}] (* A239241 *)
Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] > Count[#, ?EvenQ] &]], {n, 0, z}] (* A239242 *)
Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] >= Count[#, ?EvenQ] &]], {n, 0, z}] (* A239243 *)
(* Peter J. C. Moses, Mar 10 2014 *)
b[n_, i_, t_] := b[n, i, t] = If[n>i*(i+1)/2, 0, If[n==0, If[t>0, 1, 0], b[n, i-1, t]+If[i>n, 0, b[n-i, i-1, t+If[Mod[i, 2]==1, 1, -1]]]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Aug 30 2016, after Alois P. Heinz *)

Formula

a(n) + A239240(n) = A000009(n) for n >=1.