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A241645
Number of partitions p of n such that (number of even numbers in p) > 2*(number of odd numbers in p).
5
0, 0, 1, 0, 2, 0, 3, 0, 5, 0, 7, 0, 11, 1, 16, 4, 25, 11, 39, 26, 62, 53, 96, 97, 151, 169, 228, 280, 344, 437, 503, 669, 731, 995, 1034, 1437, 1463, 2042, 2014, 2864, 2780, 3947, 3780, 5397, 5139, 7317, 6913, 9842, 9340, 13183, 12519, 17609, 16859, 23416
OFFSET
0,5
COMMENTS
Each number in p is counted once, regardless of its multiplicity.
FORMULA
a(n) = A241644(n) - A241643(n) for n >= 0.
a(n) + A241641(n) + A241643(n) = A000041(n) for n >= 0.
EXAMPLE
a(8) counts these 5 partitions: 8, 62, 44, 422, 2222.
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] < 2 s1[p]], {n, 0, z}] (* A241641 *)
Table[Count[f[n], p_ /; s0[p] <= 2 s1[p]], {n, 0, z}] (* A241642 *)
Table[Count[f[n], p_ /; s0[p] == 2 s1[p]], {n, 0, z}] (* A241643 *)
Table[Count[f[n], p_ /; s0[p] >= 2 s1[p]], {n, 0, z}] (* A241644 *)
Table[Count[f[n], p_ /; s0[p] > 2 s1[p]], {n, 0, z}] (* A241645 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 27 2014
STATUS
approved