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Number of partitions p of n such that (number of numbers of the form 5k + 1 in p) is a part of p.
5

%I #4 May 06 2014 15:03:30

%S 0,1,1,2,3,5,7,10,14,21,28,39,51,70,92,122,158,206,265,343,432,554,

%T 695,879,1098,1373,1703,2115,2607,3218,3937,4831,5882,7175,8699,10541,

%U 12733,15358,18464,22184,26548,31774,37891,45166,53681,63743,75529,89381

%N Number of partitions p of n such that (number of numbers of the form 5k + 1 in p) is a part of p.

%C Each number in p is counted once, regardless of its multiplicity.

%e a(6) counts these 7 partitions: 51, 411, 321, 3111, 2211, 21111, 111111.

%t z = 30; f[n_] := f[n] = IntegerPartitions[n]; s[k_, p_] := Count[Mod[DeleteDuplicates[p], 5], k]

%t Table[Count[f[n], p_ /; MemberQ[p, s[0, p]]], {n, 0, z}] (* A241549 *)

%t Table[Count[f[n], p_ /; MemberQ[p, s[1, p]]], {n, 0, z}] (* A241550 *)

%t Table[Count[f[n], p_ /; MemberQ[p, s[2, p]]], {n, 0, z}] (* A241551 *)

%t Table[Count[f[n], p_ /; MemberQ[p, s[3, p]]], {n, 0, z}] (* A241552 *)

%t Table[Count[f[n], p_ /; MemberQ[p, s[4, p]]], {n, 0, z}] (* A241553 *)

%Y Cf. A241549, A241551, A241552, A241553.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Apr 26 2014