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A241655
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Number of partitions p of n such that 2*(number of even numbers in p) = (number of odd numbers in p).
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5
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0, 0, 1, 1, 3, 4, 6, 9, 12, 17, 21, 31, 37, 54, 66, 92, 114, 159, 198, 268, 335, 448, 563, 736, 921, 1190, 1485, 1892, 2340, 2953, 3636, 4534, 5542, 6861, 8333, 10226, 12347, 15052, 18079, 21907, 26168, 31537, 37526, 44987, 53307, 63653, 75156, 89369, 105204
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OFFSET
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0,5
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COMMENTS
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Each number in p is counted once, regardless of its multiplicity.
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LINKS
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FORMULA
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EXAMPLE
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a(6) counts these 6 partitions: 6, 42, 411, 222, 2211, 21111.
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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