A239873 Number of strict partitions of 2n + 1 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, 0, 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 43, 51, 61, 74, 91, 113, 144, 184, 239, 311, 407, 530, 692, 895, 1155, 1478, 1882, 2375, 2983, 3715, 4602, 5660, 6925, 8418, 10187, 12257, 14686, 17514, 20809, 24624, 29049, 34154, 40051, 46842, 54668, 63667

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 2n having 1 more even part than odd, so that there is an ordering of parts for which the even and odd parts alternate and the first and last terms are even. This sequence is nondecreasing, unlike A239871, of which it is a bisection; the other bisection is A239872.

Examples

			a(7) counts these 9 partitions of 15:  [12,1,2], [10,1,4], [10,3,2], [4,9,2], [8,1,6], [8,5,2], [8,3,4], [6,7,2], [6,5,4].
		

Links

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

Crossrefs

Cf. A239241, A239871, A239872, A239832.

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(2*n+1$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[2 n + 1]], {n, 0, 38}]  (* A239873 *)
(* 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[2n+1, 2n+1, 1]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Aug 29 2016, after Alois P. Heinz *)