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A241653
Number of partitions p of n such that 2*(number of even numbers in p) = (number of odd numbers in p).
5
1, 0, 0, 0, 0, 0, 1, 1, 4, 5, 11, 12, 24, 25, 42, 46, 70, 72, 106, 110, 156, 157, 212, 218, 291, 295, 383, 391, 516, 524, 679, 712, 931, 978, 1280, 1392, 1820, 2002, 2609, 2920, 3816, 4310, 5547, 6350, 8118, 9286, 11749, 13502, 16892, 19391, 23996, 27498
OFFSET
0,9
COMMENTS
Each number in p is counted once, regardless of its multiplicity.
FORMULA
a(n) = A241652(n) - A241651(n) for n >= 0.
a(n) + A241651(n) + A241655(n) = A000041(n) for n >= 0.
EXAMPLE
a(6) counts this single partition: 321.
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_ /; 2 s0[p] < s1[p]], {n, 0, z}] (* A241651 *)
Table[Count[f[n], p_ /; 2 s0[p] <= s1[p]], {n, 0, z}] (* A241652 *)
Table[Count[f[n], p_ /; 2 s0[p] == s1[p]], {n, 0, z}] (* A241653 *)
Table[Count[f[n], p_ /; 2 s0[p] >= s1[p]], {n, 0, z}] (* A241654 *)
Table[Count[f[n], p_ /; 2 s0[p] > s1[p]], {n, 0, z}] (* A241655 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 27 2014
STATUS
approved