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A241091 Number of partitions p of n into distinct parts such that max(p) <= 1 + 2*(number of parts of p). 3

%I #4 Apr 24 2014 10:29:28

%S 0,1,1,2,1,2,3,3,3,4,5,5,7,7,9,10,11,12,15,16,19,22,24,27,30,34,37,43,

%T 47,53,59,66,72,82,88,99,109,120,131,146,160,176,194,212,233,256,279,

%U 304,334,362,396,431,471,510,558,604,659,714,776,839,913,985

%N Number of partitions p of n into distinct parts such that max(p) <= 1 + 2*(number of parts of p).

%F a(n) = A241086(n) + A241092(n) for n >= 0.

%e a(12) counts these 5 partitions: 741, 732, 651, 642, 6321, 543, 5421.

%t z = 30; f[n_] := f[n] = Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &];

%t Table[Count[f[n], p_ /; Max[p] < 1 + 2*Length[p]], {n, 0, z}] (*A241086*)

%t Table[Count[f[n], p_ /; Max[p] <= 1 + 2*Length[p]], {n, 0, z}](*A241091*)

%t Table[Count[f[n], p_ /; Max[p] == 1 + 2*Length[p]], {n, 0, z}](*A241092*)

%t Table[Count[f[n], p_ /; Max[p] >= 1 + 2*Length[p]], {n, 0, z}](*A241089*)

%t Table[Count[f[n], p_ /; Max[p] > 1 + 2*Length[p]], {n, 0, z}] (*A241093*)

%Y Cf. A241086, A241092, A241093.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Apr 18 2014

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