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

a(n) is numerator of (A164555(n)/A027642(n) + 1/(n+1)).

1/(n+1) and Bernoulli(n,1) are autosequences in the sense that they remain the same (up to sign) under inverse binomial transform. This feature is kept for their sum, a(n)/A174342(n) = 2, 1, 1/2, 1/4, 1/6, 1/6, 1/6, 1/8, 7/90, 1/10, ...

Similar autosequences are also A000045, A001045, A113405, A000975 preceded by two zeros, and A140096.

Conjecture: the numerator of (A164555(n)/(n+1) + A027642(n)/(n+1)^2) is a(n) and the denominator of this fraction is equal to 1 if and only if n+1 is prime or 1. Cf. A309132. - Thomas Ordowski, Jul 09 2019

The "if" part of the conjecture is true: see the theorems in A309132 and A326690. The values of the numerator when n+1 is prime are A327033. - Jonathan Sondow, Aug 15 2019