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

%I #4 Apr 27 2014 10:24:08

%S 0,1,1,2,3,5,5,9,11,17,22,30,41,53,73,92,121,155,200,255,324,408,516,

%T 643,796,1009,1231,1529,1872,2317,2792,3452,4168,5073,6115,7433,8875,

%U 10741,12816,15400,18344,21923,25997,30999,36693,43412,51334,60629,71339

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

%F a(n) + A241380(n) = A000041(n) for n >= 0.

%e a(6) counts these 5 partitions: 41, 321, 2211, 21111, 111111.

%t z = 30; 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. A241377, A241378, A241379, A241380, A000041.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Apr 21 2014