A182629 Total number of largest parts in all partitions of n that contain at least two distinct parts.
0, 0, 0, 1, 2, 6, 8, 17, 23, 36, 51, 75, 95, 138, 181, 236, 310, 407, 516, 667, 840, 1062, 1344, 1678, 2080, 2589, 3212, 3942, 4851, 5937, 7246, 8824, 10724, 12971, 15705, 18895, 22749, 27296, 32734, 39083, 46668, 55553, 66086, 78389, 92937, 109857, 129850
Offset: 0
Keywords
Examples
For n = 6 the partitions of 6 are 6 ....................... all parts are equal. 5 + 1 ................... contains only one largest part. 4 + 2 ................... contains only one largest part. 4 + 1 + 1 ............... contains only one largest part. 3 + 3 ................... all parts are equal. 3 + 2 + 1 ............... contains only one largest part. 3 + 1 + 1 + 1 ........... contains only one largest part. 2 + 2 + 2 ............... all parts are equal. 2 + 2 + 1 + 1 ........... contains two largest parts. 2 + 1 + 1 + 1 + 1 ....... contains only one largest part. 1 + 1 + 1 + 1 + 1 + 1 ... all parts are equal. There are 8 largest parts, so a(6) = 8.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Programs
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Maple
b:= proc(n, i) option remember; `if`(n=i, n, 0)+ `if`(i<1, 0, b(n, i-1) +`if`(n b(n, n) -numtheory[sigma](n): seq(a(n), n=0..100); # Alois P. Heinz, Jan 17 2013
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Mathematica
b[n_, i_] := b[n, i] = If[n == i, n, 0] + If[i < 1, 0, b[n, i - 1] + If[n < i, 0, b[n - i, i]]]; a[n_] := b[n, n] - DivisorSigma[1, n]; a[0] = 0; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 06 2017, after Alois P. Heinz *)
Formula
Extensions
More terms a(13)-a(46) from David Scambler, Jul 15 2011
Comments