A220477 Total number of parts in all partitions of n with at least one distinct part.
0, 0, 2, 5, 14, 23, 46, 71, 115, 174, 263, 371, 542, 756, 1044, 1432, 1947, 2605, 3478, 4588, 6020, 7863, 10182, 13114, 16820, 21480, 27254, 34489, 43423, 54491, 68103, 84864, 105318, 130408, 160828, 197923, 242774, 297141, 362531, 441456, 536062, 649695
Offset: 1
Examples
For n = 6 ----------------------------------------------------- Partitions of 6 Value ----------------------------------------------------- 6 .......................... 0 (all parts are equal) 5 + 1 ...................... 2 4 + 2 ...................... 2 4 + 1 + 1 .................. 3 3 + 3 ...................... 0 (all parts are equal) 3 + 2 + 1 .................. 3 3 + 1 + 1 + 1 .............. 4 2 + 2 + 2 .................. 0 (all parts are equal) 2 + 2 + 1 + 1 .............. 4 2 + 1 + 1 + 1 + 1 .......... 5 1 + 1 + 1 + 1 + 1 + 1 ...... 0 (all parts are equal) ----------------------------------------------------- The sum of the values is 23 On the other hand the total number of parts of the partitions of 6 is A006128(6) = 35 and the sum of divisor of 6 is 1 + 2 + 3 + 6 = sigma(6) = A000203(6) = 12 equals the total number of parts of the partitions of 6 into equal parts. So a(6) = 35 - 12 = 23.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..1000
Programs
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Maple
b:= proc(n, i) option remember; local f, g; if n=0 or i=1 then [1, n] else f, g:= b(n, i-1), `if`(i>n, [0$2], b(n-i, i)); [f[1]+g[1], f[2]+g[2] +g[1]] fi end: a:= n-> b(n, n)[2] -numtheory[sigma](n): seq(a(n), n=1..50); # Alois P. Heinz, Jan 17 2013 -
Mathematica
a[n_] := Sum[DivisorSigma[0, k]*PartitionsP[n-k], {k, 1, n}] - DivisorSigma[1, n]; Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Oct 22 2015 *)
Formula
G.f.: Q(0)/(1-x), where Q(k)= 1 - prod(n=1..k+1, (1-x^n))/( 1 - (x^(k+1)) - x*(1- (x^(k+1)))^2*(k+2)/( x*(1- (x^(k+1)))*(k+2) - (k+1)*(1 - (x^(k+2)))*prod(n=1..k+1, (1-x^n) )/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 16 2013
Comments