cp's OEIS Frontend

zeta(4n)/zeta(2n)^2 is a rational value expressible in term of Bernoulli's numbers (A027641).

Conjecture: if an integer n > 1 is odd, then zeta(2n)/zeta(n)^2 is irrational. Cf. W. Kohnen (link) and my conjecture in A348829. - Thomas Ordowski, Jan 05 2022

Conjecture: (1 - t(n))/(1 + t(n)) = 1/2^n + 1/3^n + 1/5^n + 1/7^n + O(1/11^n), where t(n) = zeta(2n)/zeta(n)^2. Cf. A348829. - Thomas Ordowski, Nov 13 2022

Integer component of the numerator of a close variant of Euler's infinite prime product zeta(2n) = Product_{k>=1} (prime(k)^(2n))/(prime(k)^(2n)-1), namely with all minus signs changed into plus signs, as follows: zeta(4n)/zeta(2n) = Product_{k>=1} (prime(k)^(2n))/(prime(k)^(2n)+1). The transcendental component is Pi^(2n).

For a detailed account of the results, including proof and relation to the zeta function, see Links for the PDF file submitted as supporting material.

The reference to Apostol is to a discussion of the equivalence of 1) zeta(2s)/zeta(s) and 2) a related infinite prime product, that is, Product_{sigma>1} prime(n)^s/(prime(n)^s + 1), with s being a complex variable such that s = sigma + i*t where sigma and t are real (following Riemann), using a type of proof different from the one posted below involving zeta(4n)/zeta(2n). On this, see also Hardy and Wright cited below. - Leo Depuydt, Nov 22 2013, Nov 27 2013

The background of the sequence is now described in the link below to L. Depuydt, The Prime Sequence ... . - Leo Depuydt, Aug 22 2014

From Robert Israel, Aug 22 2014: (Start)

Numerator of (-1)^n*B(4*n)*4^n*(2*n)!/(B(2*n)*(4*n)!), where B(n) are the Bernoulli numbers (see A027641 and A027642).

Not the same as abs(A001067(2*n)): they differ first at n=17.

(End)

Generally, for a complex number s, w(s) = tanh(Sum_{p prime} arctanh(1/p^s)), assuming that Re(s) > 1.

Theorem. If Re(s) > 1, then w(s) = (1 - t(s))/(1 + t(s)) with t(s) = zeta(2s)/zeta(s)^2, where zeta(z) is the Riemann zeta function of z.

Proof. Einstein's formula w = (u + v)/(1 + uv) can be expanded as (1-w)/(1+w) = ((1-u)/(1+u))((1-v)/(1+v))... for any number of velocities u, v, ... Hence, by the Euler product, Product_{p prime} (1-1/p^s)/(1+1/p^s) = zeta(2s)/zeta(s)^2, qed. Note that the function f(x) = (1-x)/(1+x) is an involution.

If an integer s > 0 is even, then w(s) is rational (related to the Bernoulli numbers B_{s} and B_{2s}).

Conjecture: if an odd integer s > 1, then w(s) is irrational. Cf. W. Kohnen (link).

Note: Apery's constant zeta(3) = 1.202... is irrational.

The rational value zeta(2n)^2 / zeta(4n) = A114363(n) / A114362(n) = Product(p^4n/(((p^2n) - 1)^2)) / Product(p^4n/((p^4n) - 1)) = Product(((p^2n) + 1)/((p^2n) - 1)), where the product is over all primes.

Denote Product(((p^2n) + 1)/((p^2n) - 1)), where the product is over xxx, by F(n, xxx). For example, zeta(2n)^2 / zeta(4n) = F(n, all primes).

By definition, F(n, all primes) = F(n,primes < P) x [(P^2n) + 1]/[(P^2n) - 1]}] x F(n, primes p > P).

For small enough P, F(n, primes p > P) will be much closer to 1 than ((P^2n) + 1)/((P^2n) - 1))), so if Q is the value for which ((Q^2n) + 1)/((Q^2n) - 1) = F(n, all primes) / F(n,primes < P), Q will only be slightly less than P, so rounding Q up to the next integer (for Q < 2) and rounding up to the next odd integer (for Q >= 2) should give P. The sequence identifies the lowest P for a particular n for which such rounding fails to give P, as follows:

The sequence entry a(n) for n > 0 = the lowest prime P for which Q < P - 2, where Q is the value for which ((Q^2n) + 1)/((Q^2n) - 1) = F(n, all primes) / F(n,primes < P).

For sufficiently large n, a(n)/(2n) is bounded below, and appears to be bounded above (see A217552).

The primes up to at least A217552(n) * (2n) can therefore be reliably generated from A114363(n) / A114362(n) as follows:

Find the value of Q such that ((Q^2n) + 1)/((Q^2n) - 1) = A114363(n) / A114362(n) and round up to the nearest integer, giving 2. Then find the value of Q such that ((Q^2n) + 1)/((Q^2n) - 1) = (A114363(n) / A114362(n))/(F(n, primes <= last value found, i.e., 2)) and round up to the nearest odd integer, giving 3. Then find the value of Q such that ((Q^2n) + 1)/((Q^2n) - 1) = (A114363(n) / A114362(n))/(F(n, primes <= last value found, i.e., 3)) and round up to the nearest odd integer, and so on.

Denominator of a close variant of Euler's infinite prime product zeta(2n) = Product_{k>=1} (prime(k)^(2n))/(prime(k)^(2n)-1), namely with all minus signs changed into plus signs, as follows: zeta(4n)/zeta(2n) = Product_{k>=1} prime(k)^(2n)/(prime(k)^(2n)+1).

For a detailed account of the results in question, including proof and relation to the zeta function, see the PDF file submitted as supporting material in A231273.

The reference to Apostol below is a discussion of the equivalence of 1) zeta(2s)/zeta(s) and 2) a related infinite prime product, that is, Product_{sigma>1} prime(n)^s/(prime(n)^s + 1), with s being a complex variable such that s = sigma + i*t where sigma and t are real (following Riemann), using a type of proof different from the one posted below involving zeta(4n)/zeta(2n). - Leo Depuydt, Nov 22 2013

Denominator of B(4*n)*4^n*(2*n)!/(B(2*n)*(4*n)!) where B(n) are the Bernoulli numbers (see A027641 and A027642). - Robert Israel, Aug 22 2014

This is a rational number.

This constant does not belong to the infinite series of prime number products of the form: Product_{p>=2} (p^(2*n)-1)/(p^(2*n)+1),

which are rational numbers equal to zeta(4*n)/(zeta(2*n))^2 = A114362(n+1)/A114363(n+1).

This number has decimal period length 230:

1.81(0781476121562952243125904486251808972503617945007235890014471780028943

5600578871201157742402315484804630969609261939218523878437047756874095

5137481910274963820549927641099855282199710564399421128798842257597684

51519536903039073806).

This is a rational number.

This constant does not belong to the infinite series of prime number products of the form: Product_{p>=2} (p^(2*n)-1)/(p^(2*n)+1),

which are rational numbers equal to zeta(4*n)/zeta^2(2*n) = A114362(n+1)/A114363(n+1).

This number has decimal period length 230:

1.25(3617945007235890014471780028943560057887120115774240231548480463096960

9261939218523878437047756874095513748191027496382054992764109985528219

9710564399421128798842257597684515195369030390738060781476121562952243

12590448625180897250).

Product_{p prime} (p^(2*n)-1)/(p^(2*n)+1) are rational numbers A114362(n)/A114363(n) = zeta(4*n)/zeta(2*n)^2.

Product_{p prime} (p^(2*n+1)-1)/(p^(2*n+1)+1) = zeta(2*(2*n+1))/zeta(2*n+1)^2.