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A241638 Number of partitions p of n such that (number of even numbers in p) = (number of odd numbers in p).

%I A241638 #7 May 19 2014 10:41:52
%S A241638 1,0,0,1,1,4,3,8,6,13,11,20,17,31,34,47,56,78,103,125,167,203,281,315,
%T A241638 433,487,673,745,989,1101,1472,1623,2116,2386,3052,3430,4347,4948,
%U A241638 6168,7104,8673,10068,12210,14234,17047,20007,23671,27869,32739,38609,45010
%N A241638 Number of partitions p of n such that (number of even numbers in p) = (number of odd numbers in p).
%C A241638 Each number in p is counted once, regardless of its multiplicity.
%H A241638 Alois P. Heinz, <a href="/A241638/b241638.txt">Table of n, a(n) for n = 0..1000</a>
%F A241638 a(n) = A241637(n) - A241636(n) = A241639(n) - A241640(n) for n >= 0.
%F A241638 a(n) + A241636(n) + A241640(n) = A000041(n) for n >= 0.
%F A241638 a(n) = A242618(n,0). - _Alois P. Heinz_, May 19 2014
%e A241638 a(6) counts these 3 partitions:  411, 2211, 21111.
%t A241638 z = 30; f[n_] := f[n] = IntegerPartitions[n]; s0[p_] := Count[Mod[DeleteDuplicates[p], 2],   0];
%t A241638 s1[p_] := Count[Mod[DeleteDuplicates[p], 2], 1];
%t A241638 Table[Count[f[n], p_ /; s0[p] < s1[p]], {n, 0, z}]  (* A241636 *)
%t A241638 Table[Count[f[n], p_ /; s0[p] <= s1[p]], {n, 0, z}] (* A241637 *)
%t A241638 Table[Count[f[n], p_ /; s0[p] == s1[p]], {n, 0, z}] (* A241638 *)
%t A241638 Table[Count[f[n], p_ /; s0[p] >= s1[p]], {n, 0, z}] (* A241639 *)
%t A241638 Table[Count[f[n], p_ /; s0[p] > s1[p]], {n, 0, z}]  (* A241640 *)
%Y A241638 Cf. A241636, A241637, A241639, A241640.
%K A241638 nonn,easy
%O A241638 0,6
%A A241638 _Clark Kimberling_, Apr 27 2014