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| NAME
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allocatedNumber of partitions p of n such that 2*(number of even numbers in p) = (number of odd fornumbers Clarkin Kimberlingp).
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| DATA
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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
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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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| FORMULA
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a(n) = A241653(n) + A241655(n) for n >= 0.
a(n) + A241651(n) + A241655(n) = A000041(n) for n >= 0.
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| EXAMPLE
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a(6) counts these 7 partitions: 6, 42, 411, 321, 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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Cf. A241651, A241652, A241653, A241655.
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| KEYWORD
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allocated
nonn,easy
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| AUTHOR
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Clark Kimberling, Apr 27 2014
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| STATUS
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approved
editing
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