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A241654
Number of partitions p of n such that 2*(number of even numbers in p) = (number of odd numbers in p).
5
1, 0, 1, 1, 3, 4, 7, 10, 16, 22, 32, 43, 61, 79, 108, 138, 184, 231, 304, 378, 491, 605, 775, 954, 1212, 1485, 1868, 2283, 2856, 3477, 4315, 5246, 6473, 7839, 9613, 11618, 14167, 17054, 20688, 24827, 29984, 35847, 43073, 51337, 61425, 72939, 86905, 102871
OFFSET
0,5
COMMENTS
Each number in p is counted once, regardless of its multiplicity.
FORMULA
a(n) = A241653(n) + A241655(n) for n >= 0.
a(n) + A241651(n) + A241655(n) = A000041(n) for n >= 0.
EXAMPLE
a(6) counts these 7 partitions: 6, 42, 411, 321, 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_ /; 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