cp's OEIS Frontend

a(n)/A027642(n) (Bernoulli numbers) provide the a-sequence for the Sheffer matrix A094816 (coefficients of orthogonal Poisson-Charlier polynomials). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The corresponding z-sequence is given by the rationals A130189(n)/A130190(n).

Harvey (2008) describes a new algorithm for computing Bernoulli numbers. His method is to compute B(k) modulo p for many small primes p and then reconstruct B(k) via the Chinese Remainder Theorem. The time complexity is O(k^2 log(k)^(2+eps)). The algorithm is especially well-suited to parallelization. - Jonathan Vos Post, Jul 09 2008

Regard the Bernoulli numbers as forming a vector = B_n, and the variant starting (1, 1/2, 1/6, 0, -1/30, ...), (i.e., the first 1/2 has sign +) as forming a vector Bv_n. The relationship between the Pascal triangle matrix, B_n, and Bv_n is as follows: The binomial transform of B_n = Bv_n. B_n is unchanged when multiplied by the Pascal matrix with rows signed (+-+-, ...), i.e., (1; -1,-1; 1,2,1; ...). Bv_n is unchanged when multiplied by the Pascal matrix with columns signed (+-+-, ...), i.e., (1; 1,-1; 1,-2,1; 1,-3,3,-1; ...). - Gary W. Adamson, Jun 29 2012

The sequence of the Bernoulli numbers B_n = a(n)/A027642(n) is the inverse binomial transform of the sequence {A164555(n)/A027642(n)}, illustrated by the fact that they appear as top row and left column in A190339. - Paul Curtz, May 13 2016

Named by de Moivre (1773; "the numbers of Mr. James Bernoulli") after the Swiss mathematician Jacob Bernoulli (1655-1705). - Amiram Eldar, Oct 02 2023

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

From Gary W. Adamson & Mats Granvik, Aug 09 2008: (Start)

The von Staudt-Clausen theorem has two parts: generating denominators of the B_2n and the actual values. Both operations can be demonstrated in triangles A143343 and A080092 by following the procedures outlined in [Wikipedia - Bernoulli numbers] and summarized in A143343.

A046886(n-1) = number of terms in row n.

The same terms in A143343 may be extracted from triangle A138239.

Extract primes from even numbered rows of triangle A143343 but also include "2" as row 1. The rows are thus 1, 2, 4, 6, ..., generating denominators of B_1, B_2, B_4, ..., as well as B_1, B_2, B_4, ..., as two parts of the von Staudt-Clausen theorem.

The denominator of B_12 = 2730 = 2*3*5*7*13 = A027642(12) and A002445(6).

For example, B_12 = -691/2730 = (1 - 1/2 - 1/3 - 1/5 - 1/7 - 1/13).

The second operation is the von Staudt-Clausen representation of Bn, obtained by starting with "1" and then subtracting the reciprocals of terms in each row. (Cf. A143343 for a detailed explanation of the operations.) (End)

From von Staudt-Clausen theorem.

The decomposition of a nonzero Bernoulli number in the Staudt-Clausen format is B(n) = A000146(n) - sum_k 1/A080092(n,k) with a set of primes A080092 characterizing the right hand side.

If we multiply this equation by the product of the primes for a given n (which is in A002445), discard the left hand side, and list individually the terms associated with A000146 and each of the k, we get row n of the current triangle .

The von Staudt-Clausen decomposition of nonzero Bernoulli numbers (see A164555 and A006954) states B(0)=1, B(1) = 1/2 = 1-1/2, B(2) = 1/6 = 1-1/2-1/3, B(4) = -1/30 = 1-1/2-1/3-1/5 etc.

We consider the denominators of the fractions in these sums, one sum per row. The first term in the sums is essentially the sequence of two 1's followed by A000146; this contributes a first column to this sequence here compared with table A080092.