%I #8 May 19 2014 15:06:29
%S 0,0,0,1,1,4,5,6,9,13,17,20,26,35,43,56,67,86,105,129,158,193,232,285,
%T 350,413,507,605,740,879,1059,1274,1521,1816,2164,2577,3059,3618,4307,
%U 5103,5989,7079,8334,9797,11483,13488,15740,18469,21536,25093,29273
%N Number of partitions of n such that (number parts having multiplicity 1) is a part and (number of parts > 1) is a part.
%F a(n) + A241512(n) + A241513(n) = A241515(n) for n >= 0.
%e a(6) counts these 5 partitions: 42, 411, 321, 3111, 21111.
%t z = 30; 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, Length[p] - Count[p, 1]]], {n, 0, z}] (* A241511 *)
%t Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && MemberQ[p, Length[p] - Count[p, 1]] ], {n, 0, z}] (* A241512 *)
%t Table[Count[f[n], p_ /; MemberQ[p, u[p]] && ! MemberQ[p, Length[p] - Count[p, 1]] ], {n, 0, z}] (* A241513 *)
%t Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && ! MemberQ[p, Length[p] - Count[p, 1]] ], {n, 0, z}] (* A241514 *)
%t Table[Count[f[n], p_ /; MemberQ[p, u[p]] || MemberQ[p, Length[p] - Count[p, 1]] ], {n, 0, z}] (* A241515 *)
%Y Cf. A241506, A241512, A241513, A241514, A241515.
%K nonn,easy
%O 0,6
%A _Clark Kimberling_, Apr 24 2014