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Number of partitions of n such that (number parts having multiplicity 1) is a part or (number of 1s) is a part.
5

%I #4 Apr 29 2014 22:45:55

%S 0,1,0,1,2,5,7,10,13,20,26,36,50,70,87,123,159,205,267,347,441,572,

%T 714,899,1141,1435,1761,2218,2729,3360,4135,5088,6184,7578,9177,11131,

%U 13479,16292,19519,23556,28212,33714,40284,48049,57061,67914,80395,95143

%N Number of partitions of n such that (number parts having multiplicity 1) is a part or (number of 1s) is a part.

%F a(n) + A241509(n) = A000041(n) for n >= 0.

%e a(6) counts these 7 partitions: 51, 42, 411, 321, 3111, 2211, 21111.

%t z = 52; f[n_] := f[n] = IntegerPartitions[n]; u[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] == 1 &]]];

%t Table[Count[f[n], p_ /; MemberQ[p, u[p]] && MemberQ[p, Count[p, 1]]], {n, 0, z}] (* A241506 *)

%t Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && MemberQ[p, Count[p, 1]] ], {n, 0, z}] (* A241507 *)

%t Table[Count[f[n], p_ /; MemberQ[p, u[p]] && ! MemberQ[p, Count[p, 1]] ], {n, 0, z}] (* A241508 *)

%t Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && ! MemberQ[p, Count[p, 1]] ], {n, 0, z}] (* A241509 *)

%t Table[Count[f[n], p_ /; MemberQ[p, u[p]] || MemberQ[p, Count[p, 1]] ], {n, 0, z}] (* A241510 *)

%Y Cf. A241506, A241507, A241508, A241509.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Apr 24 2014