A217926 A sequence relating to the rational values zeta(2n)^2/zeta(4n), which are expressible in terms of Bernoulli's numbers (see comments for definition of the sequence).

7, 17, 29, 29, 41, 41, 59, 59, 71, 97, 97, 97, 101, 101, 101, 137, 137, 149, 149, 149, 179, 179, 179, 191, 191, 191, 191, 191, 227, 227, 227, 227, 227, 269, 269, 269, 269, 307, 307, 311, 311, 311, 311, 347, 347, 347, 347, 347, 347, 419, 419, 419, 419, 419

Offset: 1

Author

Roger Thompson, Oct 15 2012

Keywords

• nonn

Comments

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.

Examples

			For n = 4, A217552(n) * (2n) = 26.4417..., so primes up to at least this value can be generated. Successive Q and rounded up Q values:
   1.99029    2
   2.99331    3
   4.95780    5
   6.96977    7
  10.63524   11
  12.73590   13
  16.12527   17
  18.42182   19
  22.27250   23
  26.81206   27 (Q < 29 - 2)
so a(4) = 29.
		

Links

• Roger Thompson, Table of n, a(n) for n = 1..1403

Crossrefs

Cf. A217552, A114362, A114363.