A277790 Numerator of sum of reciprocals of proper divisors of n.
0, 1, 1, 3, 1, 11, 1, 7, 4, 17, 1, 9, 1, 23, 23, 15, 1, 19, 1, 41, 31, 35, 1, 59, 6, 41, 13, 55, 1, 71, 1, 31, 47, 53, 47, 5, 1, 59, 55, 89, 1, 95, 1, 83, 77, 71, 1, 41, 8, 46, 71, 97, 1, 119, 71, 17, 79, 89, 1, 167, 1, 95, 103, 63, 83, 13, 1, 125, 95, 143, 1, 97, 1, 113, 41, 139, 95, 167, 1, 37
Offset: 1
Examples
a(4) = 3 because 4 has 3 divisors {1,2,4} therefore 2 proper divisors {1,2} and 1/1 + 1/2 = 3/2.
0, 1, 1, 3/2, 1, 11/6, 1, 7/4, 4/3, 17/10, 1, 9/4, 1, 23/14, 23/15, 15/8, 1, 19/9, 1, 41/20, 31/21, 35/22, 1, 59/24, 6/5, 41/26, 13/9, 55/28, ...
Links
- Antti Karttunen, Table of n, a(n) for n = 1..16384
- Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
- Eric Weisstein's World of Mathematics, Restricted Divisor Function
- Index entries for sequences related to sums of divisors
Programs
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Maple
with(numtheory): P:=proc(n) local a,k; a:=divisors(n) minus {n}; numer(add(1/a[k],k=1..nops(a))); end: seq(P(i),i=1..80); # Paolo P. Lava, Oct 17 2018 -
Mathematica
Table[Numerator[DivisorSigma[-1, n] - 1/n], {n, 1, 80}] Table[Numerator[(DivisorSigma[1, n] - 1)/n], {n, 1, 80}] -
PARI
a(n) = numerator((sigma(n)-1)/n); \\ Michel Marcus, Nov 01 2016
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Python
from math import gcd from sympy import divisor_sigma def A277790(n): return (m:=divisor_sigma(n)-1)//gcd(n,m) # Chai Wah Wu, Jul 18 2022
Formula
a(n) = numerator(Sum_{d|n, d
a(n) = numerator((sigma_1(n)-1)/n).
a(p) = 1 when p is prime.
a(p^k) = (p^k - 1)/(p - 1) when p is prime.
Dirichlet g.f.: (zeta(s) - 1)*zeta(s+1) (for fraction Sum_{d|n, d