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A241639
Number of partitions p of n such that (number of even numbers in p) >= (number of odd numbers in p).
6
1, 0, 1, 1, 3, 4, 6, 9, 12, 17, 21, 31, 37, 54, 66, 91, 113, 155, 193, 254, 317, 411, 517, 649, 814, 1009, 1268, 1548, 1925, 2335, 2907, 3484, 4309, 5156, 6343, 7535, 9231, 10949, 13340, 15782, 19091, 22555, 27179, 32025, 38377, 45171, 53852, 63267, 75076
OFFSET
0,5
COMMENTS
Each number in p is counted once, regardless of its multiplicity.
LINKS
FORMULA
a(n) = A241638(n) + A241640(n) for n >= 0.
a(n) + A241636(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 6 partitions: 6, 42, 411, 222, 2211, 21111.
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