OFFSET
1,3
COMMENTS
The number of these integers is A206369(n).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
FORMULA
MATHEMATICA
f1[p_, e_] := Sum[(-1)^(e - k)*p^k, {k, 0, e}]; f2[p_, e_] := (p^(e + 1) + 1)/(p + 1); a[1] = 1; a[n_] := Module[{fct = FactorInteger[n]}, (n/2) * (Times @@ f1 @@@ fct) * If[AllTrue[fct[[;; , 2]], EvenQ], (1 + 1/(Times @@ f2 @@@ fct)), 1]]; Array[a, 100]
PROG
(PARI) a(n) = {my(f = factor(n), p, e); (n/2) * prod(i = 1, #f~, p = f[i, 1]; e = f[i, 2]; sum(k = 0, e, (-1)^(e-k) * p^k)) * if(vecsum(apply(x->x%2, f[, 2])), 1, 1 + 1/prod(i = 1, #f~, (f[i, 1]^(f[i, 2] + 1) + 1)/(f[i, 1] + 1))); }
(Python)
from math import prod
from sympy import factorint
def A390808(n):
f = factorint(n)
k = n*prod((lambda x:x[0]+int((x[1]<<1)>=p+1))(divmod(p**(e+1), p+1)) for p, e in f.items())
if any(e&1 for e in f.values()):
return k>>1
else:
m = prod((p**(e+1)+1)//(p+1) for p, e in f.items())
return k*(m+1)//(m<<1) # Chai Wah Wu, Mar 14 2026
CROSSREFS
The sum of the integers k from 1 to n such that gcd(n, k) is: A023896 (1), A119790 (prime power, A246655), A390800 (power of prime, A000961), A390801 (prime), A390802 (odd), A390803 (5-rough), A390804 (power of 2), A390805 (3-smooth), A390806 (squarefree), A390807 (cubefree), this sequence (square), A390809 (1 or 2).
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Nov 20 2025
STATUS
approved
