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%I #23 Oct 22 2015 08:35:26
%S 0,0,2,5,14,23,46,71,115,174,263,371,542,756,1044,1432,1947,2605,3478,
%T 4588,6020,7863,10182,13114,16820,21480,27254,34489,43423,54491,68103,
%U 84864,105318,130408,160828,197923,242774,297141,362531,441456,536062,649695
%N Total number of parts in all partitions of n with at least one distinct part.
%C Also total number of parts in all partitions of n minus the sum of divisors of n. Also sum of largest parts of all partitions of n minus the sum of divisors of n.
%H Alois P. Heinz, <a href="/A220477/b220477.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = A006128(n) - A000203(n).
%F 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
%e For n = 6
%e -----------------------------------------------------
%e Partitions of 6 Value
%e -----------------------------------------------------
%e 6 .......................... 0 (all parts are equal)
%e 5 + 1 ...................... 2
%e 4 + 2 ...................... 2
%e 4 + 1 + 1 .................. 3
%e 3 + 3 ...................... 0 (all parts are equal)
%e 3 + 2 + 1 .................. 3
%e 3 + 1 + 1 + 1 .............. 4
%e 2 + 2 + 2 .................. 0 (all parts are equal)
%e 2 + 2 + 1 + 1 .............. 4
%e 2 + 1 + 1 + 1 + 1 .......... 5
%e 1 + 1 + 1 + 1 + 1 + 1 ...... 0 (all parts are equal)
%e -----------------------------------------------------
%e The sum of the values is 23
%e 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.
%p b:= proc(n, i) option remember; local f, g;
%p if n=0 or i=1 then [1, n]
%p else f, g:= b(n, i-1), `if`(i>n, [0$2], b(n-i, i));
%p [f[1]+g[1], f[2]+g[2] +g[1]]
%p fi
%p end:
%p a:= n-> b(n, n)[2] -numtheory[sigma](n):
%p seq(a(n), n=1..50); # _Alois P. Heinz_, Jan 17 2013
%t 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 *)
%Y Cf. A000005, A000203, A000041, A006128, A066186, A182629, A182977, A182978.
%K nonn,easy
%O 1,3
%A _Omar E. Pol_, Jan 16 2013
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