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

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, 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, 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

Offset: 1

Author

M. F. Hasler, Nov 03 2009

Keywords

• cons
• easy
• nice
• nonn

Comments

This is a Liouville number and therefore transcendental.

Examples

			1 + 1/2^1 + 1/3^2 + 1/5^9 + 1/7^(5^9)+ ... = 1.6111116231111111111111111111111111111111...
Since 1/9 = 0.11111... and 1/5^9 = 512*10^(-9), the initial 10 digits are 1.611111623.
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....
		

Links

• J. Sondow, Exponential Factorial.
• Index entries for transcendental numbers

Crossrefs

Cf. A080219.

Programs

Mathematica

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 *)

PARI

1+1/2+1/3^2+1/5^9+1/7^5^9. /* The final dot is part of the code! */