cp's OEIS Frontend

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).

%I A217926 #41 Feb 28 2022 02:05:12
%S A217926 7,17,29,29,41,41,59,59,71,97,97,97,101,101,101,137,137,149,149,149,
%T A217926 179,179,179,191,191,191,191,191,227,227,227,227,227,269,269,269,269,
%U A217926 307,307,311,311,311,311,347,347,347,347,347,347,419,419,419,419,419
%N 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).
%C A217926 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.
%C A217926 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).
%C A217926 By definition, F(n, all primes) = F(n,primes < P) x [(P^2n) + 1]/[(P^2n) - 1]}] x F(n, primes p > P).
%C A217926 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:
%C A217926 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).
%C A217926 For sufficiently large n, a(n)/(2n) is bounded below, and appears to be bounded above (see A217552).
%C A217926 The primes up to at least A217552(n) * (2n) can therefore be reliably generated from A114363(n) / A114362(n) as follows:
%C A217926 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.
%H A217926 Roger Thompson, <a href="/A217926/b217926.txt">Table of n, a(n) for n = 1..1403</a>
%e A217926 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:
%e A217926    1.99029    2
%e A217926    2.99331    3
%e A217926    4.95780    5
%e A217926    6.96977    7
%e A217926   10.63524   11
%e A217926   12.73590   13
%e A217926   16.12527   17
%e A217926   18.42182   19
%e A217926   22.27250   23
%e A217926   26.81206   27 (Q < 29 - 2)
%e A217926 so a(4) = 29.
%Y A217926 Cf. A217552, A114362, A114363.
%K A217926 nonn
%O A217926 1,1
%A A217926 _Roger Thompson_, Oct 15 2012