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%I #9 Jun 19 2021 03:01:04
%S 7,8,2,5,2,7,9,8,5,3,2,5,3,8,4,2,3,4,5,7,6,6,8,8,4,7,4,2,8,3,7,8,4,0,
%T 7,6,8,0,5,8,0,0,9,7,9,6,5,2,5,9,9,1,5,9,9,9,2,6,4,7,2,0,8,7,8,2,6,0,
%U 7,3,7,6,7,1,9,9,0,0,3,5,2,2,9,2,7,1,6
%N Decimal expansion of Sum_{k>=1} 1/(2^k*zeta(2*k)).
%C Equals the alternating sum of a sequence of real numbers c(k) (see the Formula section). The Riemann Hypothesis is equivalent to c(k) ~ O(k^(-3/4+eps)) for all eps>0 (Báez-Duarte, 2005).
%C The sum without the alternating signs is Sum_{k>=0} c(k) = 1/zeta(0) = -2.
%H Luis Báez-Duarte, <a href="https://doi.org/10.1155/IJMMS.2005.3527">A sequential Riesz-like criterion for the Riemann hypothesis</a>, International Journal of Mathematics and Mathematical Sciences, Vol. 2005, No. 21 (2005), pp. 3527-3537.
%H Jerzy Cisło and Marek Wolf, <a href="https://arxiv.org/abs/math/0607782">Equivalence of Riesz and Baez-Duarte criterion for the Riemann Hypothesis</a>, arXiv:math/0607782 [math.NT], 2006.
%H Jerzy Cisło and Marek Wolf, <a href="https://doi.org/10.1063/1.3043867">Criteria equivalent to the Riemann Hypothesis</a>, AIP Conference Proceedings, Vol. 1079, No. 1. (2008), pp. 268-273; <a href="https://arxiv.org/abs/0808.0640">arXiv preprint</a>, arXiv:0808.0640 [math.NT], 2008.
%H Jerzy Cisło and Marek Wolf, <a href="https://arxiv.org/abs/0807.2971">On the Riesz and Baez-Duarte criteria for the Riemann Hypothesis</a>, arXiv:0807.2971 [math.NT], 2008.
%H Mark W. Coffey, <a href="https://arxiv.org/abs/math-ph/0608050">On the coefficients of the Baez-Duarte criterion for the Riemann hypothesis and their extensions</a>, arXiv:math-ph/0608050, 2006.
%F Equals Sum_{k>=0} (-1)^k * c(k), where c(k) = Sum_{n>=1} mu(n)*(1-1/n^2)^k/n^2 = Sum_{j=0..k} (-1)^j * binomial(k,j)/zeta(2*j+2), where mu is the Möbius function (A008683).
%F Equals 1 + Integral_{x>=2} (1 - 1/2^floor(x/2)) * zeta'(x)/zeta(x) dx.
%e 0.78252798532538423457668847428378407680580097965259...
%p evalf(Sum(1/(2^k*Zeta(2*k)), k = 1..infinity), 120); # _Vaclav Kotesovec_, Jun 19 2021
%t RealDigits[Sum[1/(2^k*Zeta[2*k]), {k,1,1000}], 10, 100][[1]]
%o (PARI) suminf(k=1, 1/(2^k*zeta(2*k)))
%Y Cf. A008683.
%K nonn,cons
%O 0,1
%A _Amiram Eldar_, Jun 18 2021