A296074 Sum of deficiencies of the proper divisors of n.

0, 1, 1, 2, 1, 4, 1, 3, 3, 6, 1, 5, 1, 8, 7, 4, 1, 9, 1, 9, 9, 12, 1, 2, 5, 14, 8, 13, 1, 16, 1, 5, 13, 18, 11, 3, 1, 20, 15, 8, 1, 24, 1, 21, 18, 24, 1, -9, 7, 27, 19, 25, 1, 20, 15, 14, 21, 30, 1, -1, 1, 32, 24, 6, 17, 40, 1, 33, 25, 40, 1, -27, 1, 38, 32, 37, 17, 48, 1, -1, 22, 42, 1, 9, 21, 44, 31, 26, 1, 18, 19, 45, 33, 48, 23, -36, 1, 53, 36, 33

Offset: 1

Author

Antti Karttunen, Dec 04 2017

Keywords

• sign
• easy

Examples

			For n = 6, whose proper divisors are 1, 2, 3, their deficiencies are 1, 1, 2, thus a(6) = 1+1+2 = 4.
For n = 12, whose proper divisors are 1, 2, 3, 4, 6, their deficiencies are 1, 1, 2, 1, 0, thus a(12) = 1+1+2+1+0 = 5.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A033879.

Cf. also A294886, A294887, A294888, A294889, A293438 (product of).

Programs

Mathematica

f1[p_, e_] := (p^(e+1)-1)/(p-1); f2[p_, e_] := (p*(p^(e+1)-1) - (p-1)*(e+1))/(p-1)^2; a[1] = 0; a[n_] := Module[{f = FactorInteger[n]}, 3 * Times @@ f1 @@@ f - Times @@ f2 @@@ f - 2*n]; Array[a, 100] (* Amiram Eldar, Dec 04 2023 *)

PARI

A033879(n) = ((2*n)-sigma(n));
A296074(n) = sumdiv(n,d,(dA033879(d));

Formula

a(n) = Sum_{d|n, dA033879(d).

a(n) = A296075(n) - A033879(n).

Sum_{k=1..n} a(k) ~ (Pi^2/4 - Pi^4/72 - 1) * n^2. - Amiram Eldar, Dec 04 2023