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Number of partitions of n 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.
8

%I #11 Apr 01 2014 05:04:48

%S 0,0,1,0,1,1,1,2,2,4,3,7,6,11,11,17,19,27,31,41,51,62,79,95,121,142,

%T 182,212,269,314,393,459,570,665,816,958,1160,1364,1639,1928,2297,

%U 2706,3200,3768,4434,5212,6105,7170,8361,9799,11396,13322,15450,18022

%N Number of partitions of n 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.

%C 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.

%e The three partitions counted by a(10) are [10], [4,1,2,1,2], and [2,3,2,1,2].

%t p[n_] := p[n] = Select[IntegerPartitions[n], Count[#, _?OddQ] == -1 + Count[#, _?EvenQ] &]; t = Table[p[n], {n, 0, 10}]

%t TableForm[t] (* shows the partitions *)

%t Table[Length[p[n]], {n, 0, 30}] (* A239832 *)

%t (* _Peter J. C. Moses_, Mar 10 2014 *)

%Y Cf. A239833, A239835, A045931, A239871.

%Y Column k=-1 of A240009.

%K nonn,easy

%O 0,8

%A _Clark Kimberling_, Mar 29 2014