|
%I #6 Apr 24 2014 10:29:41
%S 0,0,0,1,0,0,1,1,1,1,1,1,2,2,3,3,3,3,4,4,5,7,7,8,9,10,10,12,13,15,17,
%T 19,21,25,26,29,32,35,38,42,46,51,57,62,69,76,83,90,100,107,117,127,
%U 139,150,165,178,195,212,231,250,273,294,319,346,373,402
%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.
%F a(n) = A241091(n) - A241086(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, A241091, A241093, A000009.
%K nonn,easy
%O 0,13
%A _Clark Kimberling_, Apr 18 2014
|
