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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

With the signs of A359738, the rational sequence reflects the identity B(z)^2 = (z + 1)*B(z) - z*B'(z), that goes back to Euler, where B(z) = z/(1 - e^(-z)) is the e.g.f. of the Bernoulli numbers with B(1) = 1/2. - Peter Luschny, Jan 23 2023

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

An autosequence is a sequence whose inverse binomial transform is the sequence signed. In integers, the oldest example is Fibonacci A000045. In fractions, A164555/A027642 is the son of 1/n via the Akiyama-Tanigawa algorithm; grandson is (A174110/A174111) = 1/2, 2/3, 1/2, 2/15, ...; see A164020. See A174341/A174342. All are from the same family.

The denominators are given in A285068.

In general the numbers B(d;n) = d^n*B(n), for n >= 0, have e.g.f. d*x/(exp(d*x) - 1). They are also the exponential convolution of the generalized Bernoulli numbers B[d,a](n), obtained from the generalized Stirling2 numbers S2[d,a], with the sequence {(-a)^n}_{n>=0}. See a comment in A157817 for the B[4,1] and B[4,3] examples.

These numbers B(d;n) and their polynomials B(d;n,x) = Sum_{m=0..n} binomial(n, m)*B(d;n-m)*x^m are used in the generalized so-called Faulhaber formula for the sums of powers of arithmetic progressions defined by SP(d,a;n,m) := Sum_{j=0..m} (a + d*j)^n = Sum_{k=0..n} binomial(n, k)*a^(n-k)*d^k*SP(k,m) with SP(k,m) = SP(1,0;k,m), n >= 0, m >= 0, and 0^0 := 1.

The Faulhaber formula is: SP(d,a;n,m) = (1/(d*(n+1)))*[B(d;n+1,x = a+d*(m+1)) - B(d;n+1,x = d) - B(d;n+1,x = a) + B(d;n+1,x=0) + d^(n+1)*[n=0]]. Here [n=0] is the Kronecker delta_{n,0} symbol: 1 if n=0 and 0 otherwise.

A simpler version of the Faulhaber formula is for a=0: SP(d,0;0,m) = m+1 and SP(d,0;n,m) = d^n*(1/(n+1))*(B(n+1, x = m+1) - B(n+1, x=1)) for n >= 1, and for a an integer >= 1: Sum_{k=0..n} binomial(n, k)*a^(n-k) * d^k * (1/(k+1)) * (B(k+1, x=m+1) - B(k+1, x=1)). Here B(n, x) = B(1;n,x) are the usual Bernoulli polynomials from A196838/A196839 or A053382/A053383.

Except for the first column, the n-th prime number appears in every A006093(n)-th row, beginning at the A000040(n)-th row, in the n-th column.

If n != 1, also the numerator of 1 - Bernoulli(n). The denominators are in A027642.

(There are no common factors to be canceled in the fractions.)

The numerators of 1 - Bernoulli(n) start 0, 3, 5,1, 31, ... and differ only at n=1 from this sequence.

E.g.f. for the rationals r(n) = a(n)/A027642(n) = 1 - A164555(n)/A027642(n): exp(x)*(1 - x/(exp(x) - 1)). - Wolfdieter Lang, Aug 07 2017

Compare the sequence for example with A120083.