cp's OEIS Frontend

It seems that the numerator of F(n) is the numerator of (B(n-1) + 1/n), where B(k) is the k-th Bernoulli number; if so, for n > 2, the numerator of F(n) is A174341(n-1). How to prove it?

Conjecture: for n > 1, a(n) = 1 if and only if n is prime.

Is this conjecture equivalent to the Agoh-Giuga conjecture?

Theorem 1. If p is prime, then a(p) = 1. Proof. a(2) = 1, so let p be an odd prime. By the von Staudt-Clausen theorem, if k is even, then B(k) = A(k) - Sum_{prime q, q-1 | k} 1/q, where A(k) is an integer and the sum is over all primes q such that q-1 divides k. Thus B(k) = N(k)/D(k) with D(k) = Product_{prime q, q-1 | k} q. Now let k = p-1. Then N(p-1)/D(p-1) = B(p-1) = A(p-1) - 1/p - Sum_{prime q < p, q-1 | p-1} 1/q (*). Add 1/p to both sides of (*) and multiply by p*D(p-1) to get p*N(p-1) + D(p-1) = p*D(p-1)*(A(p-1) - Sum_{prime q < p, q-1 | p-1} 1/q) (**). Now p | D(p-1), so p^2 | p*D(p-1) in (**). The denominators on the right side of (**) are all of the form q < p. Therefore, p^2 divides both sides of (**). Hence F(p) = N(p-1)/p + D(p-1)/p^2 is an integer, so a(p) = 1. - Jonathan Sondow, Jul 14 2019

Conjecture: composite numbers n such that a(n) is squarefree are only the Carmichael numbers A002997. Cf. A309235. - Thomas Ordowski, Jul 15 2019

Conjecture checked up to n = 101101. - Amiram Eldar, Jul 16 2019

Theorem 2. If n is a prime or a Carmichael number, then a(n) = A326690(n) = denominator of (Sum_{prime p | n} 1/p - 1/n). The proof is a generalization of that of Theorem 1. (Note that Theorem 2 implies Theorem 1, since if n is prime, then (Sum_{prime p | n} 1/p - 1/n) = 1/n - 1/n = 0/1, so a(p) = A326690(n) = 1.) For n a prime or a Carmichael number, an application of Theorem 2 is computing a(n) without calculating Bernoulli(n-1) which may be huge; see A309268 and A326690. - Jonathan Sondow, Jul 19 2019

The values of F(n) when n is prime are A327033. - Jonathan Sondow, Aug 16 2019

As in A166062, the offset is rather arbitrary.

The sequence contains numbers like 210 which are not in A006954.

One could also consider dividing by the largest prime divisor of A027642 instead of A089026, which yields 1, 1, 2, 1, 6, 1, 6, 1, 6, 1, 6, 1, 210, 1, 2, 1, 30, 1, 42, 1, 30, ... as an alternative version.

These are the Clausen numbers based on the proper divisors of n whereas the classical Clausen numbers A160014 are based on all divisors of n. (The proper divisors are the divisors of n that are less than n.) - Peter Luschny, Aug 20 2022