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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

%I #4 May 03 2014 16:53:47

%S 1,0,1,1,3,4,7,10,16,22,32,43,61,79,108,138,184,231,304,378,491,605,

%T 775,954,1212,1485,1868,2283,2856,3477,4315,5246,6473,7839,9613,11618,

%U 14167,17054,20688,24827,29984,35847,43073,51337,61425,72939,86905,102871

%N Number of partitions p of n such that 2*(number of even numbers in p) = (number of odd numbers in p).

%C Each number in p is counted once, regardless of its multiplicity.

%F a(n) = A241653(n) + A241655(n) for n >= 0.

%F a(n) + A241651(n) + A241655(n) = A000041(n) for n >= 0.

%e a(6) counts these 7 partitions: 6, 42, 411, 321, 222, 2211, 21111.

%t z = 30; f[n_] := f[n] = IntegerPartitions[n]; s0[p_] := Count[Mod[DeleteDuplicates[p], 2], 0];

%t s1[p_] := Count[Mod[DeleteDuplicates[p], 2], 1];

%t Table[Count[f[n], p_ /; 2 s0[p] < s1[p]], {n, 0, z}] (* A241651 *)

%t Table[Count[f[n], p_ /; 2 s0[p] <= s1[p]], {n, 0, z}] (* A241652 *)

%t Table[Count[f[n], p_ /; 2 s0[p] == s1[p]], {n, 0, z}] (* A241653 *)

%t Table[Count[f[n], p_ /; 2 s0[p] >= s1[p]], {n, 0, z}] (* A241654 *)

%t Table[Count[f[n], p_ /; 2 s0[p] > s1[p]], {n, 0, z}] (* A241655 *)

%Y Cf. A241651, A241652, A241653, A241655.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Apr 27 2014

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Last modified April 23 18:16 EDT 2024. Contains 371916 sequences. (Running on oeis4.)