A296075 Sum of deficiencies of divisors of n.

1, 2, 3, 3, 5, 4, 7, 4, 8, 8, 11, 1, 13, 12, 13, 5, 17, 6, 19, 7, 19, 20, 23, -10, 24, 24, 22, 13, 29, 4, 31, 6, 31, 32, 33, -16, 37, 36, 37, -2, 41, 12, 43, 25, 30, 44, 47, -37, 48, 34, 49, 31, 53, 8, 53, 6, 55, 56, 59, -49, 61, 60, 46, 7, 63, 28, 67, 43, 67, 36, 71, -78, 73, 72, 58, 49, 75, 36, 79, -27, 63, 80, 83, -47, 83

Offset: 1

Author

Antti Karttunen, Dec 04 2017

Keywords

• sign
• easy

Comments

a(n)=0 for n in A066218. Are 1 and 12 the only solutions to a(n)=1? - Robert Israel, Dec 04 2017

Examples

			For n = 6, whose divisors are 1, 2, 3, 6, their deficiencies are 1, 1, 2, 0, thus a(6) = 1 + 1 + 2 + 0 = 4.
For n = 24, whose divisors are 1, 2, 3, 4, 6, 8, 12, 24, their deficiencies are 1, 1, 2, 1, 0, 1, -4, -12, thus a(24) = 1 + 1 + 2 + 1 + 0 + 1 + -4 + -12 = -10.
		

Links

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

Crossrefs

Cf. A000203, A033879, A066218, A296074.

Cf. also A007429, A187793, A187794, A187795.

Programs

Maple

f:= n -> add(2*t-numtheory:-sigma(t), t=numtheory:-divisors(n)):
map(f, [$1..100]); # Robert Israel, Dec 04 2017

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] = 1; a[n_] := Module[{f = FactorInteger[n]}, 2 * Times @@ f1 @@@ f - Times @@ f2 @@@ f]; Array[a, 100] (* Amiram Eldar, Dec 04 2023 *)

PARI

A033879(n) = ((2*n)-sigma(n));
A296075(n) = sumdiv(n,d,A033879(d));

Formula

a(n) = Sum_{d|n} A033879(d).

a(n) = A296074(n) + A033879(n).

If m and n are coprime, a(m*n) = 2*a(m)*A000203(n)+2*a(n)*A000203(m)-a(m)*a(n)-2*A000203(m)*A000203(n). - Robert Israel, Dec 04 2017

a(n) = 2*A000203(n) - A007429(n). - Ridouane Oudra, Jul 29 2019

Sum_{k=1..n} a(k) ~ (Pi^2/6 - Pi^4/72) * n^2. - Amiram Eldar, Dec 04 2023