OFFSET
1,2
LINKS
Jinyuan Wang, Table of n, a(n) for n = 1..1000
FORMULA
From Amiram Eldar, Jun 30 2020: (Start)
Equals 1/2 + Sum_{k>=1} (-1)^(k+1)/(2^k-1)
Equals Sum_{k>=1} (mu(k) - (-1)^k)/(2^k-1), where mu is the Möbius function (A008683).
Equals (1 + A179951)/2. (End)
Equals (1/2) + Sum_{n >= 1} (-1)^(n+1)*((4^n + 1)/(4^n - 1))*(1/2)^(n^2). The first 18 terms of the series gives the constant correct to more than 100 decimal places. - Peter Bala, Jan 28 2022
EXAMPLE
1.2644997803484442091913...
MAPLE
evalf((1/2) + add( (-1)^(n+1)*((4^n + 1)/(4^n - 1))*(1/2)^(n^2), n = 1..18), 100); # Peter Bala, Jan 28 2022
MATHEMATICA
s = Sum[1/(2^k + 1), {k, 0, Infinity}]
r = N[Re[s], 200]
RealDigits[r][[1]]
PROG
(PARI) suminf(k=0, 1/(2^k+1)) \\ Michel Marcus, Jan 15 2019
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Clark Kimberling, Jan 15 2019
STATUS
approved