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A241636
Number of partitions p of n such that (number of even numbers in p) < (number of odd numbers in p).
6
0, 1, 1, 2, 2, 3, 5, 6, 10, 13, 21, 25, 40, 47, 69, 85, 118, 142, 192, 236, 310, 381, 485, 606, 761, 949, 1168, 1462, 1793, 2230, 2697, 3358, 4040, 4987, 5967, 7348, 8746, 10688, 12675, 15403, 18247, 22028, 25995, 31236, 36798, 43963, 51706, 61487, 72197
OFFSET
0,4
COMMENTS
Each number in p is counted once, regardless of its multiplicity.
LINKS
FORMULA
a(n) + A241638(n) = A241637(n) for n >= 0.
a(n) + A241638(n) + A241640(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 5 partitions: 51, 33, 321, 3111, 111111.
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