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A240312 Number of partitions p of n such that (maximal multiplicity of the parts of p) = (maximal part of p). 5

%I #6 Dec 25 2023 17:38:13

%S 1,1,0,0,2,1,2,0,2,3,5,5,9,7,11,11,18,15,28,27,41,43,62,64,91,96,127,

%T 140,184,200,260,287,365,410,511,573,717,803,985,1120,1359,1538,1859,

%U 2106,2522,2870,3407,3872,4586,5207,6128,6976,8167,9284,10844,12321

%N Number of partitions p of n such that (maximal multiplicity of the parts of p) = (maximal part of p).

%F a(n) = A240311(n) - A240310(n) for n >= 0.

%F a(n) + A240310(n) + A240314(n) = A000041(n) for n >= 0.

%e a(6) counts these 2 partitions: 3111, 2211.

%t z = 60; f[n_] := f[n] = IntegerPartitions[n]; m[p_] := Max[Map[Length, Split[p]]] (* maximal multiplicity *)

%t Table[Count[f[n], p_ /; m[p] < Max[p]], {n, 0, z}] (* A240310 *)

%t Table[Count[f[n], p_ /; m[p] <= Max[p]], {n, 0, z}] (* A240311 *)

%t Table[Count[f[n], p_ /; m[p] == Max[p]], {n, 0, z}] (* A240312 *)

%t Table[Count[f[n], p_ /; m[p] >= Max[p]], {n, 0, z}] (* A240313 *)

%t Table[Count[f[n], p_ /; m[p] > Max[p]], {n, 0, z}] (* A240314 *)

%Y Cf. A240310, A240311, A240313, A240314, A000041.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Apr 05 2014

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Last modified August 27 15:02 EDT 2024. Contains 375470 sequences. (Running on oeis4.)