A241091 -id:A241091

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A241092

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

+10
3

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, 19, 21, 25, 26, 29, 32, 35, 38, 42, 46, 51, 57, 62, 69, 76, 83, 90, 100, 107, 117, 127, 139, 150, 165, 178, 195, 212, 231, 250, 273, 294, 319, 346, 373, 402

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OFFSET

0,13

LINKS

Table of n, a(n) for n=0..65.

FORMULA

a(n) + A241086(n) + A241093(n) = A000009(n) for n >= 1.

a(n) = A241091(n) - A241086(n) for n >= 0.

EXAMPLE

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

MATHEMATICA

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

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

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

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

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

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

CROSSREFS

Cf. A241086, A241091, A241093, A000009.

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Apr 18 2014

STATUS

approved

A241093

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