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 A239950 Number of partitions of n such that (number of distinct parts) = least part. 5
 0, 1, 1, 1, 1, 2, 2, 3, 4, 4, 6, 6, 9, 8, 14, 11, 19, 18, 25, 24, 37, 31, 50, 46, 61, 64, 86, 79, 112, 115, 136, 149, 190, 184, 239, 255, 293, 329, 382, 408, 489, 531, 595, 675, 772, 827, 952, 1066, 1176, 1320, 1468, 1627, 1827, 2030, 2219, 2493, 2769, 3053 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,6 COMMENTS Also for n>0 the number of partitions of n such that (number of distinct parts) = multiplicity of the greatest part (by conjugation of the partition table). - Joerg Arndt, Apr 28 2014 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..1000 FORMULA A239948(n) + a(n) + A239951(n) = A000041(n) for n >= 0. EXAMPLE a(8) counts these 4 partitions :  62, 422, 332, 11111111. MAPLE b:= proc(n, i, d) option remember; `if`(min(i, n) b(n\$2, 0): seq(a(n), n=0..60);  # Alois P. Heinz, Apr 02 2014 MATHEMATICA z = 50; d[p_] := d[p] = Length[DeleteDuplicates[p]]; f[n_] := f[n] = IntegerPartitions[n]; Table[Count[f[n], p_ /; d[p] < Min[p]], {n, 0, z}]  (*A239948*) Table[Count[f[n], p_ /; d[p] <= Min[p]], {n, 0, z}] (*A239949*) Table[Count[f[n], p_ /; d[p] == Min[p]], {n, 0, z}] (*A239950*) Table[Count[f[n], p_ /; d[p] > Min[p]], {n, 0, z}]  (*A239951*) Table[Count[f[n], p_ /; d[p] >= Min[p]], {n, 0, z}] (*A239952*) b[n_, i_, d_] := b[n, i, d] = If[Min[i, n]

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Last modified August 11 03:22 EDT 2020. Contains 336421 sequences. (Running on oeis4.)