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Number of partitions p of n such that 2*(number of even numbers in p) = (number of odd numbers in p).
5

%I #4 May 03 2014 16:53:55

%S 0,0,1,1,3,4,6,9,12,17,21,31,37,54,66,92,114,159,198,268,335,448,563,

%T 736,921,1190,1485,1892,2340,2953,3636,4534,5542,6861,8333,10226,

%U 12347,15052,18079,21907,26168,31537,37526,44987,53307,63653,75156,89369,105204

%N Number of partitions p of n such that 2*(number of even numbers in p) = (number of odd numbers in p).

%C Each number in p is counted once, regardless of its multiplicity.

%F a(n) = A241654(n) - A241653(n) for n >= 0.

%F a(n) + A241651(n) + A241653(n) = A000041(n) for n >= 0.

%e a(6) counts these 6 partitions: 6, 42, 411, 222, 2211, 21111.

%t z = 30; f[n_] := f[n] = IntegerPartitions[n]; s0[p_] := Count[Mod[DeleteDuplicates[p], 2], 0];

%t s1[p_] := Count[Mod[DeleteDuplicates[p], 2], 1];

%t Table[Count[f[n], p_ /; 2 s0[p] < s1[p]], {n, 0, z}] (* A241651 *)

%t Table[Count[f[n], p_ /; 2 s0[p] <= s1[p]], {n, 0, z}] (* A241652 *)

%t Table[Count[f[n], p_ /; 2 s0[p] == s1[p]], {n, 0, z}] (* A241653 *)

%t Table[Count[f[n], p_ /; 2 s0[p] >= s1[p]], {n, 0, z}] (* A241654 *)

%t Table[Count[f[n], p_ /; 2 s0[p] > s1[p]], {n, 0, z}] (* A241655 *)

%Y Cf. A241651, A241652, A241653, A241654.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Apr 27 2014