A239871 Number of strict partitions of n having 1 more even part than odd, so that there is at least one ordering of the parts in which the even and odd parts alternate, and the first and last terms are even.

0, 0, 1, 0, 1, 0, 1, 1, 1, 2, 1, 4, 1, 6, 1, 9, 2, 12, 3, 16, 6, 20, 10, 25, 17, 30, 26, 36, 40, 43, 57, 51, 81, 61, 110, 74, 148, 91, 193, 113, 250, 144, 316, 184, 397, 239, 491, 311, 603, 407, 732, 530, 885, 692, 1061, 895, 1268, 1155, 1508, 1478, 1790

Offset: 0

Author

Clark Kimberling, Mar 29 2014

Keywords

• nonn
• easy

Comments

Let c(n) be the number of strict partitions (that is, every part has multiplicity 1) of n having 1 more odd part than even, so that there is an ordering of parts for which the odd and even parts alternate and the first and last terms are odd. Then c(n) = a(n+1) for n >= 0.

Examples

			a(11) counts these 4 partitions:  812, 614, 632, 452.
		

Links

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

Crossrefs

Cf. A239241, A239872, A239873, A239832.

Column k=-1 of A240021.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n>i*(i+1)/2 or
      abs(t)>n, 0, `if`(n=0, 1, b(n, i-1, t)+
      `if`(i>n, 0, b(n-i, i-1, t+(2*irem(i, 2)-1)))))
    end:
a:= n-> b(n$2, 1):
seq(a(n), n=0..80);  # Alois P. Heinz, Apr 02 2014

Mathematica

d[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &];
p[n_] := p[n] = Select[d[n], Count[#, ?OddQ] == -1 + Count[#, ?EvenQ] &]; t =  Table[p[n], {n, 0, 20}]
TableForm[t] (* shows the partitions *)
u = Table[Length[p[n]], {n, 0, 70}]  (* A239871 *)
(* Peter J. C. Moses, Mar 10 2014 *)
b[n_, i_, t_] := b[n, i, t] = If[n > i*(i + 1)/2 || Abs[t] > n, 0, If[n == 0, 1, b[n, i - 1, t] + If[i > n, 0, b[n - i, i - 1, t + (2*Mod[i, 2] - 1)]]]]; a[n_] := b[n, n, 1]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Nov 16 2015, after Alois P. Heinz *)

Formula

a(n) = [x^n y^(-1)] Product_{i>=1} 1+x^i*y^(2*(i mod 2)-1). - Alois P. Heinz, Apr 03 2014