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Number of partitions of n such that the number of parts is a part and the number of distinct parts is a part.
5

%I #5 Apr 27 2014 10:23:29

%S 0,1,0,1,0,1,2,2,3,4,5,6,8,11,12,16,20,27,31,40,49,59,72,95,110,133,

%T 164,196,237,289,351,410,502,595,704,843,1009,1193,1422,1658,1983,

%U 2332,2744,3204,3796,4459,5189,6083,7116,8292,9677,11222,13041,15235

%N Number of partitions of n such that the number of parts is a part and the number of distinct parts is a part.

%F a(n) + A241378(n) + A241379(n) = A241381(n) for n >= 0.

%e a(9) counts these 4 partitions: 72, 531, 432, 4311.

%t z = 70; f[n_] := f[n] = IntegerPartitions[n]; d[p_] := [p] = Length[DeleteDuplicates[p]];

%t Table[Count[f[n], p_ /; MemberQ[p, Length[p]] && MemberQ[p, d[p]]], {n, 0, z}] (* A241377 *)

%t Table[Count[f[n], p_ /; ! MemberQ[p, Length[p]] && MemberQ[p, d[p]]], {n, 0, z}] (* A241378 *)

%t Table[Count[f[n], p_ /; MemberQ[p, Length[p]] && ! MemberQ[p, d[p]]], {n, 0, z}] (* A241379 *)

%t Table[Count[f[n], p_ /; ! MemberQ[p, Length[p]] && ! MemberQ[p, d[p]]], {n, 0, z}] (* A241380 *)

%t Table[Count[f[n], p_ /; MemberQ[p, Length[p]] || MemberQ[p, d[p]]], {n, 0, z}] (* A241381 *)

%Y Cf. A241378, A241379, A241380, A241381.

%K nonn,easy

%O 0,7

%A _Clark Kimberling_, Apr 21 2014