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A167155 Exponential primorial constant Sum_{k>=0} 1/A140319(k).

%I A167155 #15 Feb 16 2025 08:33:11
%S A167155 1,6,1,1,1,1,1,6,2,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%T A167155 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%U A167155 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
%N A167155 Exponential primorial constant Sum_{k>=0} 1/A140319(k).
%C A167155 This is a Liouville number and therefore transcendental.
%H A167155 J. Sondow, <a href="https://mathworld.wolfram.com/ExponentialFactorial.html">Exponential Factorial</a>.
%H A167155 <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%e A167155 1 + 1/2^1 + 1/3^2 + 1/5^9 + 1/7^(5^9)+ ... = 1.6111116231111111111111111111111111111111...
%e A167155 Since 1/9 = 0.11111... and 1/5^9 = 512*10^(-9), the initial 10 digits are 1.611111623.
%e A167155 Since 1/A140319(4) = 1/7^1953125 = 7.7731519...*10^(-1650583), these digits are followed by a string of 1650573 "1"s, then followed by digits 8884263011....
%t A167155 Clear[ep, s]; ep[0] = 1; ep[n_] := Prime[n]^ep[n-1]; s[n_] := s[n] = RealDigits[Sum[1/ep[k], {k, 0, n}], 10, 105] // First; s[n=1]; While[s[n] != s[n-1], n++]; s[n] (* _Jean-François Alcover_, Feb 13 2013 *)
%o A167155 (PARI) 1+1/2+1/3^2+1/5^9+1/7^5^9. /* The final dot is part of the code! */
%Y A167155 Cf. A080219.
%K A167155 cons,easy,nice,nonn
%O A167155 1,2
%A A167155 _M. F. Hasler_, Nov 03 2009