%I #55 Nov 02 2024 17:34:40
%S 3,0,6,8,5,2,8,1,9,4,4,0,0,5,4,6,9,0,5,8,2,7,6,7,8,7,8,5,4,1,8,2,3,4,
%T 3,1,9,2,4,4,9,9,8,6,5,6,3,9,7,4,4,7,4,5,8,7,9,3,1,9,9,9,0,5,0,6,6,0,
%U 6,3,7,8,0,3,0,3,0,5,2,8,4,3,9,4,1,3,6,6,7,3,0,0,3,5,8,1,3,1,2,4,5,7,9,9,8,5
%N Decimal expansion of 1 - log(2).
%C Fraction of numbers which are sqrt-smooth, see A048098 and A063539. - _Charles R Greathouse IV_, Jul 14 2014
%C Asymptotic survival probability in the 100 prisoners problem. - _Alois P. Heinz_, Jul 08 2022
%D Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.6.3, pp. 43-44.
%H Peter Bala, <a href="/A244009/a244009.pdf">A continued fraction expansion for the constant 1 - log(2)</a>
%H Donald E. Knuth and Luis Trabb Pardo, <a href="http://dx.doi.org/10.1016/0304-3975(76)90050-5">Analysis of a simple factorization algorithm</a>, Theoretical Computer Science 3:3 (1976), pp. 321-348.
%H The Ramanujan Machine, <a href="http://www.ramanujanmachine.com/">Using algorithms to discover new mathematics</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/100_prisoners_problem">100 prisoners problem</a>
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F Equals Sum_{k>=0} 1/(2*k*(2*k+1)) = A239354 + 1/4 = A188859/2.
%F From _Amiram Eldar_, Aug 07 2020: (Start)
%F Equals Sum_{k>=1} 1/(k*(k+1)*2^k) = Sum_{k>=2} 1/A100381(k).
%F Equals Sum_{k>=2} (-1)^k * zeta(k)/2^k.
%F Equals Integral_{x=1..oo} 1/(x^2 + x^3) dx. (End)
%F Equals log(e/2) = log(A019739) = -log(2/e) = -log(A135002). - _Wolfdieter Lang_, Mar 04 2022
%F Equals lim_{n->oo} A024168(n)/n!. - _Alois P. Heinz_, Jul 08 2022
%F Equals 1/(4 - 4/(7 - 12/(10 - ... - 2*n*(n-1)/((3*n+1) - ...)))) (an equivalent continued fraction for 1 - log(2) was conjectured by the Ramanujan machine). - _Peter Bala_, Mar 04 2024
%F Equals Sum_{k>=1} zeta(2*k)/((2*k + 1)*2^(2*k-1)) (see Finch). - _Stefano Spezia_, Nov 02 2024
%e 0.30685281944005469058276787854...
%p f:= sum(1/(2*k*(2*k+1)), k=1..infinity):
%p s:= convert(evalf(f, 140), string):
%p seq(parse(s[i+1]), i=1..106); # _Alois P. Heinz_, Jun 17 2014
%t RealDigits[1-Log[2],10,120][[1]] (* _Harvey P. Dale_, Sep 23 2016 *)
%o (PARI) 1-log(2) \\ _Charles R Greathouse IV_, Jul 14 2014
%Y Cf. A002162, A019739, A024168, A100381, A135002, A188859.
%Y Essentially the same digits as A239354.
%K nonn,cons
%O 0,1
%A _Franklin T. Adams-Watters_, Jun 17 2014