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A241511
Number of partitions of n such that (number parts having multiplicity 1) is a part and (number of parts > 1) is a part.
5
0, 0, 0, 1, 1, 4, 5, 6, 9, 13, 17, 20, 26, 35, 43, 56, 67, 86, 105, 129, 158, 193, 232, 285, 350, 413, 507, 605, 740, 879, 1059, 1274, 1521, 1816, 2164, 2577, 3059, 3618, 4307, 5103, 5989, 7079, 8334, 9797, 11483, 13488, 15740, 18469, 21536, 25093, 29273
OFFSET
0,6
FORMULA
a(n) + A241512(n) + A241513(n) = A241515(n) for n >= 0.
EXAMPLE
a(6) counts these 5 partitions: 42, 411, 321, 3111, 21111.
MATHEMATICA
z = 30; f[n_] := f[n] = IntegerPartitions[n]; u[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] == 1 &]]];
Table[Count[f[n], p_ /; MemberQ[p, u[p]] && MemberQ[p, Length[p] - Count[p, 1]]], {n, 0, z}] (* A241511 *)
Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && MemberQ[p, Length[p] - Count[p, 1]] ], {n, 0, z}] (* A241512 *)
Table[Count[f[n], p_ /; MemberQ[p, u[p]] && ! MemberQ[p, Length[p] - Count[p, 1]] ], {n, 0, z}] (* A241513 *)
Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && ! MemberQ[p, Length[p] - Count[p, 1]] ], {n, 0, z}] (* A241514 *)
Table[Count[f[n], p_ /; MemberQ[p, u[p]] || MemberQ[p, Length[p] - Count[p, 1]] ], {n, 0, z}] (* A241515 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 24 2014
STATUS
approved