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Number of partitions p of n such that (sum of parts with multiplicity 1) > (sum of all other parts).
5

%I #6 Apr 06 2020 20:29:34

%S 0,1,1,2,2,4,5,8,10,17,20,28,35,55,66,87,106,153,183,243,293,401,477,

%T 608,723,984,1162,1458,1720,2245,2624,3301,3864,4963,5775,7108,8246,

%U 10508,12153,14834,17125,21442,24651,30028,34477,42599,48778,58742,67091

%N Number of partitions p of n such that (sum of parts with multiplicity 1) > (sum of all other parts).

%F a(n) + A240449(n) = A000041(n) for n >= 0.

%e a(6) counts these 5 partitions: 6, 51, 42, 411, 321.

%t z = 30; p[n_] := p[n] = IntegerPartitions[n]; f[p_] := f[p] = First[Transpose[p]];

%t ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] < n &], {n, 0, z}]] (* shows the partitions *)

%t Map[Length, t] (* A240448 *)

%t ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] <= n &], {n, 0, z}]] (* shows the partitions *)

%t Map[Length, t] (* A240449 *)

%t ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] == n &], {n, 0, z}]] (* shows the partitions *)

%t Map[Length, t] (* A240447 with alternating 0's *)

%t ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] > n &], {n, 0, z}]] (* shows the partitions *)

%t Map[Length, t] (* A240451 *)

%t ColumnForm[t = Table[Select[p[n], 2 Total[f[Select[#, Last[#] == 1 &] /. {} -> {{0, 0}}]] &[Tally[#]] >= n &], {n, 0, z}]] (* shows the partitions *)

%t Map[Length, t] (* A240452 *)

%t (* _Peter J. C. Moses_, Apr 02 2014 *)

%Y Cf. A240448, A240447, A240449, A240452, A000041.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Apr 05 2014