%I #13 Sep 25 2022 04:28:13
%S 1,9,7,4,2,5,5,0,2,3,0,6,4,6,5,2,5,9,3,3,7,8,3,2,6,7,5,5,1,0,9,9,6,9,
%T 5,3,6,1,8,9,0,5,8,8,9,5,6,8,5,7,0,8,3,3,0,6,1,2,5,6,0,1,8,1,5,7,5,2,
%U 1,4,0,6,4,0,3,0,9,7,7,5,0,6,9,5,3,3,1,7,9,8,1,2,9,3,0,6,6,7,7,6
%N Decimal expansion of the constant that satisfies (2^x+1)*(3^x+1)^2/(2^x*(3^(2*x)+1)) = zeta(x)/zeta(2x).
%C See the Defant link for more explanations on this constant.
%H Colin Defant, <a href="https://arxiv.org/abs/1507.02654">Ranges of Unitary Divisor Functions</a>, arXiv:1507.02654 [math.NT], 2015-2018.
%H Colin Defant, <a href="http://math.colgate.edu/~integers/s15/s15.Abstract.html">Ranges of Unitary Divisor Functions</a>, Integers, 18 (2018), #A15.
%e 1.97425502306465259337832675510996953618905889568570833061256...
%t RealDigits[x /. FindRoot[(2^x+1)*(3^x+1)^2/(2^x*(3^(2*x)+1)) == Zeta[x]/Zeta[2*x], {x, 3/2}, WorkingPrecision -> 120]][[1]] (* _Amiram Eldar_, Sep 25 2022 *)
%o (PARI) solve(x=1.5, 2, (2^x+1)*(3^x+1)^2*zeta(2*x) - (2^x*(3^(2*x)+1)*zeta(x)))
%Y Cf. A294795.
%K cons,nonn
%O 1,2
%A _Michel Marcus_, Aug 26 2018