login
A241644
Number of partitions p of n such that (number of even numbers in p) >= 2*(number of odd numbers in p).
5
1, 0, 1, 0, 2, 0, 3, 1, 6, 4, 10, 11, 20, 23, 32, 44, 56, 76, 86, 124, 136, 193, 199, 293, 297, 430, 422, 619, 609, 884, 855, 1246, 1217, 1742, 1708, 2438, 2423, 3393, 3415, 4717, 4845, 6558, 6828, 9097, 9653, 12585, 13549, 17379, 18987, 23897, 26420, 32712
OFFSET
0,5
COMMENTS
Each number in p is counted once, regardless of its multiplicity.
FORMULA
a(n) = A241643(n) + A241645(n) for n >= 0.
a(n) + A241641(n) = A000041(n) for n >= 0.
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] < 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