cp's OEIS Frontend

It appears that, except for irregular primes belonging to A094095, such as a(5) = 103, a(8) = 37 and a(26) = 59, all regular prime a(n) = p divide the corresponding numerators of the Bernoulli numbers B(A090997(n)) with indices of the form 2*k*p^2, where k > 0 is an integer. - Alexander Adamchuk, Aug 19 2006

For each m in the current sequence, the smallest prime whose cube divides the numerator of the Bernoulli number B(m) is listed in A122271.

The current sequence is a subset of A090997, which are numbers m such that the numerator of the Bernoulli number B(m) is divisible by a square.

A subset of the current sequence is A122272, which are numbers m such that the numerator of the Bernoulli number B(m) is divisible by a fourth power.

Conjecture: For all regular primes p > 3 and integers k > 0, the numerator of the Bernoulli number B(2*p^k) is divisible by p^k. Moreover, for all regular primes p > 3 and integers k > 0, m = 2*p^k is the smallest index such that the numerator of the Bernoulli number B(m) is divisible by p^k. Also, for all regular primes p > 3 and integers k > 0, all m such that the numerator of the Bernoulli number B(m) is divisible by p^k are of the form m = 2*s*p^k, where s > 0 is an integer.

For each m in the current sequence, the smallest prime p such that p^4 divides the numerator of the Bernoulli number B(m) is listed in A122273.

Note that the numerator of B(6250) is divisible by 5^5.

The current sequence is a subsequence of A122270, which are the numbers m such that the numerator of the Bernoulli number B(m) is divisible by a cube.

Sequence A122270 itself is a subsequence of A090997, which are the numbers m such that the numerator of the Bernoulli number B(m) is divisible by a square. [Edited by Petros Hadjicostas, May 12 2020]

The numbers m in A122272 are such that the numerator of the Bernoulli number B(m) is divisible by p^4, where p is a prime. For m = 6250, we have that the numerator of B(6250) is divisible by 5^5.

Equivalently, n not coprime to the numerator of 2^(2n-1)*Bernoulli(2n)/(2n)! (see Lekraj Beedassy comment in A046988).

Conjecture 1: for n>=1, a(n) is identical to 2*A072823(n+1).

Conjecture 2: The corresponding GCDs are powers of 2.

Verified for n <= 10000, e.g.,

GCD = 2 for 14, 22, 26, 28, 30, 38, 42, 44, 46, 50, 52, 54, 56, 58, ...

GCD = 4 for 60, 92, 108, 116, 120, 124, 156, 172, 180, 184, 188, ...

GCD = 8 for 248, 376, 440, 472, 488, 496, 504, 632, 696, 728, 744, ...

GCD = 16 for 1008, 1520, 1776, 1904, 1968, 2000, 2016, 2032, 2544, ...

GCD = 32 for 4064, 6112, 7136, 7648, 7904, 8032, 8096, 8128, 8160

Taking GCDs vertically, column 1 = "14, 60, 248, 1008, 4064, ..." appears to be essentially the same as A171499 and A131262; (ii) column 2 = "22, 92, 376, 1520, 6112, ..." appears to be essentially the same as A010036.

From Chris Boyd, Jan 25 2016: (Start)

Determining whether n is a term of this sequence can be approached by considering odd and even factors separately, and exploiting the fact that numerator(zeta(2n)/(Pi^(2n))) = numerator(2^(2n-2)*N_2n/(D_2n*(2n)!)), where N_2n and D_2n are odd coprime integers such that Bernoulli(2n) = N_2n/2D_2n.

Case 1: odd factors. n is a term if it has an odd prime factor p (necessarily irregular) that divides N_2n at a higher multiplicity than it divides (2n)!. No such factor p of N_2n up to 2n = 10000 is of sufficient multiplicity, and the apparent scarcity of squared and higher power factors of N_2n values (see A090997) suggests that no such p is likely to exist.

Case 2: even factors. An even n is a term if 2 divides 2^(2n-2) at a higher multiplicity than it divides (2n)!. The multiplicity of 2 in 2^(2n-2) is 2n-2, and in (2n)! is 2n minus the number of 1's in the binary expansion of 2n (see A005187). Qualifying n values are therefore those where the number of 1's in the binary expansion of 2n is greater than 2. Except for its first term, A072823 comprises integers with three or more 1-bits in their binary expansion. It follows that for m > 1, 2*A072823(m) values belong to this sequence.

In summary, this sequence is essentially a supersequence of 2*A072823. Conjectures 1 and 2 are true if there are no irregular odd primes p that divide n and the numerator of Bernoulli(2n)/(2n)!. (End)