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

%I #6 Apr 24 2014 10:29:48

%S 0,0,0,0,1,1,1,2,3,4,5,7,8,11,13,17,21,26,31,38,45,54,65,77,92,108,

%T 128,149,175,203,237,274,318,366,424,486,559,640,733,836,953,1084,

%U 1232,1398,1583,1792,2025,2286,2576,2902,3262,3666,4111,4610,5160,5774

%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) + A241093(n) = A000009(n) for n >= 1.

%e a(12) counts these 8 partitions: {12}, {11,1}, {10,2}, {9,3}, 9,2,1}, {8,4}, {8,3,1}, {7,5}.

%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, A241091, A241092, A000009.

%K nonn,easy

%O 0,8

%A _Clark Kimberling_, Apr 18 2014