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A240449
Number of partitions p of n such that (sum of parts with multiplicity 1) <= (sum of all other parts).
5
1, 0, 1, 1, 3, 3, 6, 7, 12, 13, 22, 28, 42, 46, 69, 89, 125, 144, 202, 247, 334, 391, 525, 647, 852, 974, 1274, 1552, 1998, 2320, 2980, 3541, 4485, 5180, 6535, 7775, 9731, 11129, 13862, 16351, 20213, 23141, 28523, 33233, 40698, 46535, 56780, 66012, 80182
OFFSET
0,5
FORMULA
a(n) + A240451(n) = A000041(n) for n >= 0.
EXAMPLE
a(6) counts these 5 partitions: 33, 3111, 222, 2211, 21111, 111111.
MATHEMATICA
z = 30; p[n_] := p[n] = IntegerPartitions[n]; f[p_] := f[p] = First[Transpose[p]];
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] < n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t] (* A240448 *)
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] <= n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t] (* A240449 *)
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] == n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t] (* A240447 with alternating 0's *)
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] > n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t] (* A240451 *)
ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] >= n &], {n, 0, z}]] (* shows the partitions *)
Map[Length, t] (* A240452 *)
(* Peter J. C. Moses, Apr 02 2014 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 05 2014
STATUS
approved