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A240180
Number of partitions of n such that (least part) = (multiplicity of greatest part).
3
0, 1, 0, 1, 3, 3, 5, 7, 12, 16, 24, 30, 45, 57, 81, 104, 143, 179, 243, 304, 399, 504, 650, 809, 1039, 1286, 1622, 2006, 2508, 3077, 3822, 4666, 5747, 6995, 8552, 10353, 12603, 15189, 18371, 22071, 26570, 31785, 38104, 45419, 54213, 64426, 76596, 90710
OFFSET
0,5
COMMENTS
Also the number of partitions p of n such that min(p) = min(conjugate(p)). Example:a(7) counts these 7 partitions: 61, 511, 421, 4111, 3211, 31111, 211111, of which the respective conjugates are 211111, 31111, 3211, 4111, 421, 511, 61. - Clark Kimberling, Apr 11 2014
FORMULA
a(n) = A240179(n) - A240178(n), for n >= 0.
a(n) + 2*A240178(n) = A000041(n) for n >= 0.
EXAMPLE
a(6) counts these 5 partitions: 51, 411, 321, 3111, 21111.
MATHEMATICA
z = 60; f[n_] := f[n] = IntegerPartitions[n];
Table[Count[f[n], p_ /; Min[p] < Count[p, Max[p]]], {n, 0, z}] (* A240178 *)
Table[Count[f[n], p_ /; Min[p] <= Count[p, Max[p]]], {n, 0, z}] (* A240179 *)
Table[Count[f[n], p_ /; Min[p] == Count[p, Max[p]]], {n, 0, z}] (* A240180 *)
Table[Count[f[n], p_ /; Min[p] > Count[p, Max[p]]], {n, 0, z}] (* A240178, n>0 *)
Table[Count[f[n], p_ /; Min[p] >= Count[p, Max[p]]], {n, 0, z}] (* A240179, n>0 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 02 2014
STATUS
approved