OFFSET
0,1
COMMENTS
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
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..158
Winfried Kohnen, Transcendence conjectures about periods of modular forms and rational structures on spaces of modular forms, Proceedings of the Indian Academy of Sciences-Mathematical Sciences, Vol. 99, No. 3 (1989), pp. 231-233.
Herbert Wilf, Problem 11068, The American Mathematical Monthly, Vol. 111, No. 3 (2004), p. 259; Think Rationally, Solution to Problem 11068 by Kenneth E. Schilling, ibid., Vol. 112, No. 9 (2005), pp. 844-845.
FORMULA
Product_{p primes} (p^{2n}-1)/(p^{2n}+1) = zeta(4n)/zeta(2n)^2.
For n > 0, a(n) = Numerator((D(n) - N(n)) / (D(n) + N(n))), where N(n) = A348829(n) and D(n) = A348830(n). See my comments and formulas in A348829. - Thomas Ordowski, Jan 05 2022
From Amiram Eldar, Mar 04 2023: (Start)
a(n)/A114363(n) = -2*B(4*n)/(binomial(4*n,*2n)*B(2*n)) = -2*(A027641(4*n)/A027642(4*n))/(A000984(2*n)*A027641(2*n)/A027642(2*n)), for n >= 1, where B(n) is the n-th Bernoulli number.
A114363(n)/a(n) = Sum_{x in Q+} 1/f(x)^(2*n), for n >= 1, where Q+ is the set of the positive rational numbers, and if x = k/m in lowest terms, then f(x) = k*m (Wilf, 2004). (End)
EXAMPLE
2/1, 2/5, 6/7, 691/715, 7234/7293, 523833/524875, 3545461365/3547206349, ...
MATHEMATICA
a[n_] := Numerator[Zeta[4*n]/Zeta[2*n]^2]; a[0] = 2; Array[a, 14, 0] (* Amiram Eldar, Mar 04 2023 *)
PROG
(PARI) z(n)=bernfrac(2*n)*(-1)^(n - 1)*2^(2*n-1)/(2*n)!;
a(n)=if(n<1, 2, numerator(z(2*n)/z(n)^2))
CROSSREFS
KEYWORD
frac,nonn
AUTHOR
Benoit Cloitre, Feb 09 2006; corrected Feb 22 2006
STATUS
approved