OFFSET
0,8
COMMENTS
Let c(n) be the number of partitions of n having 1 more odd part than even, so that there is an ordering of parts for which the even and odd parts alternate and the first and last terms are odd. Then c(n) = a(n+1) for n >= 0.
EXAMPLE
The three partitions counted by a(10) are [10], [4,1,2,1,2], and [2,3,2,1,2].
MATHEMATICA
p[n_] := p[n] = Select[IntegerPartitions[n], Count[#, _?OddQ] == -1 + Count[#, _?EvenQ] &]; t = Table[p[n], {n, 0, 10}]
TableForm[t] (* shows the partitions *)
Table[Length[p[n]], {n, 0, 30}] (* A239832 *)
(* Peter J. C. Moses, Mar 10 2014 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 29 2014
STATUS
approved