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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).
38
1, 0, 0, 1, 1, 4, 3, 8, 6, 13, 11, 20, 17, 31, 34, 47, 56, 78, 103, 125, 167, 203, 281, 315, 433, 487, 673, 745, 989, 1101, 1472, 1623, 2116, 2386, 3052, 3430, 4347, 4948, 6168, 7104, 8673, 10068, 12210, 14234, 17047, 20007, 23671, 27869, 32739, 38609, 45010
OFFSET
0,6
COMMENTS
Each number in p is counted once, regardless of its multiplicity.
LINKS
FORMULA
a(n) = A241637(n) - A241636(n) = A241639(n) - A241640(n) for n >= 0.
a(n) + A241636(n) + A241640(n) = A000041(n) for n >= 0.
a(n) = A242618(n,0). - Alois P. Heinz, May 19 2014
EXAMPLE
a(6) counts these 3 partitions: 411, 2211, 21111.
MATHEMATICA
z = 30; f[n_] := f[n] = IntegerPartitions[n]; s0[p_] := Count[Mod[DeleteDuplicates[p], 2], 0];
s1[p_] := Count[Mod[DeleteDuplicates[p], 2], 1];
Table[Count[f[n], p_ /; s0[p] < s1[p]], {n, 0, z}] (* A241636 *)
Table[Count[f[n], p_ /; s0[p] <= s1[p]], {n, 0, z}] (* A241637 *)
Table[Count[f[n], p_ /; s0[p] == s1[p]], {n, 0, z}] (* A241638 *)
Table[Count[f[n], p_ /; s0[p] >= s1[p]], {n, 0, z}] (* A241639 *)
Table[Count[f[n], p_ /; s0[p] > s1[p]], {n, 0, z}] (* A241640 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 27 2014
STATUS
approved