login
A241413
Number of partitions p of n such that the number of numbers having multiplicity 1 in p is a part of p.
6
0, 1, 0, 1, 1, 4, 5, 8, 10, 17, 21, 29, 38, 59, 68, 100, 124, 170, 214, 288, 351, 470, 576, 743, 921, 1176, 1430, 1816, 2214, 2753, 3364, 4176, 5015, 6215, 7478, 9120, 10966, 13351, 15916, 19301, 22982, 27618, 32846, 39354, 46515, 55570, 65598, 77842, 91730
OFFSET
0,6
EXAMPLE
a(6) counts these 5 partitions: 42, 411, 321, 3111, 21111; e.g., 411 is counted because 1 part of 411 has multiplicity 1, and 1 is a part of 411.
MATHEMATICA
z = 30; f[n_] := f[n] = IntegerPartitions[n]; u[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] == &]]]; e[q_] := Length[DeleteDuplicates[Select[q, Count[q, #] > 1 &]]]
Table[Count[f[n], p_ /; MemberQ[p, u[p]]], {n, 0, z}] (* A241413 *)
Table[Count[f[n], p_ /; MemberQ[p, u[p]] && MemberQ[p, e[p]]], {n, 0, z}] (* A241414 *)
Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && MemberQ[p, e[p]] ], {n, 0, z}] (* A241415 *)
Table[Count[f[n], p_ /; MemberQ[p, u[p]] && ! MemberQ[p, e[p]] ], {n, 0, z}] (* A241416 *)
Table[Count[f[n], p_ /; ! MemberQ[p, u[p]] && ! MemberQ[p, e[p]] ], {n, 0, z}] (* A241417 *)
Table[Count[f[n], p_ /; MemberQ[p, u[p]] || MemberQ[p, e[p]] ], {n, 0, z}] (* A239737 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 23 2014
STATUS
approved