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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

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.

Squares of Bernoulli number numerators (apart from the sign flipped at a(1)).

The associated denominators of the squared Bernoulli numbers are in A172282.

Formally, these are the numerators of a sequence of fractions defined by alternating A164555(n)/A027642(n) with A027641(n)/A027642(n),

which apart from the third term duplicates the Bernoulli numbers.

Essentially a duplication of the entries of A027641.

Essentially the same as A176144. (The signs of the third and fourth entry are swapped.)

This refers to a shuffling of the "original" Bernoulli numbers and the Bernoulli numbers in opposite order compared to the composition discussed in A176150.

The inverse binomial transform of the shuffle in A176150 was 1,0, -1/2, 0, 13/6, -20/3. The shuffling here would yield an inverse binomial transform 1, 0, -3/2, 4, -47/6, 40/3, -21, 95/3 etc.

The difference between the corresponding elements of these two binomial transforms element by element is 0, 0, 1, -4, 10, -20, 35, -56, 84, -120, 165, -220,..., a signed variant of A000292.

The denominator triangle b(n,m) is given in A285865.

a(n,m)/b(n,m) = B(2;n,m) is the d = 2 instance of the fractional d-family of triangles B(d;n,m) = binomial(n,m)*d^(n-m)*B(n-m), for d >= 1. They are the coefficient triangles of generalized Bernoulli polynomials PB(d;n,x) = Sum_{m=0..n} B(d;n,m)*x^m for n >= 0.

{PB(d;n,x)}{n>=0} has e.g.f. EB(d;x,z) := Sum{n>=0} PB(d;n,x)*z^n = d*z*exp(x*z)/(exp(d*z)-1). B(d;n,m) is a Sheffer triangle of the Appell type for each d, denoted by (d*z/(exp(d*z - 1)), z).

PB(d;n,x) gives a (trivial) generalization of the Bernoulli polynomials with coefficients given in A196838/A196839 (rising powers of x), and this is PB(1;n,x).

The polynomials PB(d;n,x) appear in the generalized Faulhaber formula for sums of powers of arithmetic progressions SP(n,m) := Sum_{j=0..m} (a + d*j)^n, n >= 0, m >= 0, d >= 1, a = 0 for d = 1 and a from the smallest positive restricted residue system modulo d >= 2. For this Faulhaber formula see a comment in A285863, where they are named B(d;n,x).

The row sums of the rational triangle B(2;n,m) give A157779(n)/A141459(n). The alternating row sums are given in A285866/A141459(n).

c(n) starts with: 0, 1, 3/2, 19/6, 19/3,3 79/30, 379/15, 3539/70, 3539/35, 42461/210, 42461/105, ...

The corresponding denominator is A172031 (also denominator of rational sequence defined in A172030).

It appears that A172030/A172031 - A172032/A172031 = 0, 0, 1, 2, 4, 8, 16, ... that is A131577 prepended with 0.

Reduced a(n)/c(n) = 0, 0, 1, 1/2, 5/3, 13/6, 19/5, 179/30, 1028/105, 1103/70, 893/35, 2889/70, 15445/231, 249787/2310,... .

After the first Bernoulli numbers we consider the same transform applied to the second Bernoulli numbers A164555(n)/A027642(n). Hence reduced b(n)/c(n) = 0, 0, 1, 3/2, 8/3, 25/6, 34/5, 329/30, 1868/105, 2013/70, 1628/35, 5269/70, 28150/231, 455377/2310, ....

Conjecture: (b(n)-a(n))/c(n) = 0, 0, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ..., that is two 0 followed by A000045.

This conjecture is confirmed up to 100 terms. [Jean-François Alcover, Jul 19 2013]