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A241652
Number of partitions p of n such that 2*(number of even numbers in p) <= (number of odd numbers in p).
5
1, 1, 1, 2, 2, 3, 5, 6, 10, 13, 21, 25, 40, 47, 69, 84, 117, 138, 187, 222, 292, 344, 439, 519, 654, 768, 951, 1118, 1378, 1612, 1968, 2308, 2807, 3282, 3977, 4657, 5630, 6585, 7936, 9278, 11170, 13046, 15648, 18274, 21868, 25481, 30402, 35385, 42069, 48875
OFFSET
0,4
COMMENTS
Each number in p is counted once, regardless of its multiplicity.
FORMULA
a(n) = A241651(n) + A241653(n) for n >= 0.
a(n) + A241655(n) = A000041(n) for n >= 0.
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_ /; 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