OFFSET
0,1
COMMENTS
For x = e^(-1/2), the largest prime factor of a random integer n is equally likely to be above or below n^x. - Charles R Greathouse IV, May 25 2009
FORMULA
Equals Sum_{k>=0} (-1)^k/(2^k * k!) = Sum_{k>=0} (-1)^k/A000165(k). - Amiram Eldar, Aug 15 2020
From Peter Bala, Jan 16 2022; (Start)
Equals 16*Sum_{n >= 0} (-1)^n*n^2/((4*n^2 - 1)*(4*n^2 - 9)*(2^n)*n!).
Equals 8*Sum_{n >= 0} (-1)^n/(p(n)*p(n+1)*(2^n)*n!), where p(n) = 4*n^2 + 8*n + 1.
Equals 48*Sum_{n >= 0} (-1)^n/(q(n)*q(n+1)*(2^n)*n!), where q(n) = 8*n^3 + 36*n^2 + 34*n + 1. (End)
EXAMPLE
0.6065306597126334...
MATHEMATICA
RealDigits[E^-(1/2), 10, 120][[1]] (* Harvey P. Dale, Jul 23 2012 *)
PROG
(PARI) exp(-.5) \\ Charles R Greathouse IV, Oct 02 2022
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
Mohammad K. Azarian, Apr 22 2004
STATUS
approved