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A241640
Number of partitions p of n such that (number of even numbers in p) > (number of odd numbers in p).
6
0, 0, 1, 0, 2, 0, 3, 1, 6, 4, 10, 11, 20, 23, 32, 44, 57, 77, 90, 129, 150, 208, 236, 334, 381, 522, 595, 803, 936, 1234, 1435, 1861, 2193, 2770, 3291, 4105, 4884, 6001, 7172, 8678, 10418, 12487, 14969, 17791, 21330, 25164, 30181, 35398, 42337, 49463, 59057
OFFSET
0,5
COMMENTS
Each number in p is counted once, regardless of its multiplicity.
LINKS
FORMULA
a(n) = A241639(n) - A241638(n) for n >= 0.
a(n) + A241636(n) + A241638(n) = A000041(n) for n >= 0.
a(n) = Sum_{k<0} A242618(n,k). - Alois P. Heinz, May 19 2014
EXAMPLE
a(6) counts these 3 partitions: 6, 42, 222.
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