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A345144 Product_{p primes, k>=1} ((p^(k+1) - 1)/(p^(k+1) - p))^(1/p^k).

%I A345144 #17 Jun 10 2021 05:05:57
%S A345144 1,5,6,1,5,9,6,8,4,6,9,3,1,0,2,4,1,6,4,3,2,6,9,6,7,8,8,9,1,4,4,5,5,5,
%T A345144 6,4,4,3,6,4,7,3,7,6,4,6,8,2,2,2,3,2,1,6,9,9,4,5,8,6,6,4,5,7,0,9,6,8,
%U A345144 3,5,7,8,4,9,4,9,0,9,5,3,9,8,8,9,4,2,4,4,3,0,1,0,8,6,8,0,9,1,0,3,2,1,4,3,7
%N A345144 Product_{p primes, k>=1} ((p^(k+1) - 1)/(p^(k+1) - p))^(1/p^k).
%H A345144 Ramanujan's Papers, <a href="https://web.archive.org/web/20200124035942/http://ramanujan.sirinudi.org/Volumes/published/ram17.html">Some formulas in the analytic theory of numbers</a>, Messenger of Mathematics, XLV, 1916, 81-84, Formula (20), constant c.
%F A345144 Equals exp(1) * lim_{n->infinity} (A066780(n)^(1/n)) / n.
%e A345144 1.561596846931024164326967889144555644364737646822232169945866457...
%t A345144 $MaxExtraPrecision = 1000; m = 500; prod = 1; Do[Clear[f]; f[p_] := ((p^(k + 1) - 1)/(p^(k + 1) - p))^(1/p^k); cc = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x, m + 1]]; prod *= f[2]*Exp[N[Sum[Indexed[cc, n]*(PrimeZetaP[n] - 1/2^n), {n, 2, m}], 100]]; Print[prod], {k, 1, 200}]
%Y A345144 Cf. A001088, A066843, A066780, A124175.
%K A345144 nonn,cons
%O A345144 1,2
%A A345144 _Vaclav Kotesovec_, Jun 09 2021