OFFSET
1,3
COMMENTS
a(k) is the sum of totatives of k modulo the sum of divisors of k.
If p is an odd prime, a(p) = (p+3)/2 and a(p^2) = (p-1)/2.
If p is a prime == 5 (mod 6), a(2*p) = 2.
If p is a prime == 1 (mod 6), a(2*p) = 2*p+4.
Are 2, 8 and 9 the only solutions to a(k) = 1?
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
EXAMPLE
a(6) = 6 because the sum of totatives of 6 is 1+5 = 6, the sum of divisors of 6 is 1+2+3+6 = 12, and 6 mod 12 = 6.
MAPLE
f:= n -> n*numtheory:-phi(n)/2 mod numtheory:-sigma(n):
map(f, [$1..100]);
MATHEMATICA
Array[Mod[# EulerPhi[#]/2 + Boole[# == 1]/2, DivisorSigma[1, #]] &, 80] (* Michael De Vlieger, Feb 23 2021 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
J. M. Bergot and Robert Israel, Feb 23 2021
STATUS
approved