A239872 Number of strict partitions of 2n 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, 1, 1, 1, 1, 1, 1, 1, 2, 3, 6, 10, 17, 26, 40, 57, 81, 110, 148, 193, 250, 316, 397, 491, 603, 732, 885, 1061, 1268, 1508, 1790, 2120, 2510, 2970, 3517, 4170, 4950, 5887, 7013, 8371, 10005, 11979, 14353, 17217, 20654, 24785, 29725, 35637, 42672, 51046, 60962

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 A239873.

Examples

			a(9) counts these 3 partitions of 18:  [18], [8,3,4,1,2], [6,5,4,1,2].
		

Links

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

Crossrefs

Cf. A239241, A239871, A239873, A239832.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n>i*(i+1)/2 or
      abs(t)-n>0, 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$2, 1):
seq(a(n), n=0..60);  # Alois P. Heinz, Apr 01 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]], {n, 0, 40}]  (* A239872 *)
(* 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, 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[2*n, 2*n, 1]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Oct 28 2015, after Alois P. Heinz *)