%I #22 Jan 30 2022 15:57:14

%S 1,5,2,8,9,9,9,5,6,0,6,9,6,8,8,8,4,1,8,3,8,2,6,3,9,4,9,4,5,1,0,9,9,6,

%T 9,6,5,1,1,5,3,9,3,9,9,7,7,1,5,0,5,1,2,5,3,1,3,2,4,7,5,9,2,0,5,3,1,7,

%U 5,1,3,5,9,5,3,2,0,1,4,1,7,0,1,2,3,8,0,8,8,6,4,3,0,5,7,0,7,9,7,0,2,2,2,7,0

%N Decimal expansion of Sum_{k has exactly two bits equal to 1 in base 2} 1/k.

%C Obviously for k > 0 in base 2 having no bit equal to 1 the sum is 0 and for 1 bit equal to 1 the sum is 2.

%H Robert Baillie, <a href="https://arxiv.org/abs/0806.4410">Summing The Curious Series Of Kempner and Irwin</a>, arXiv:0806.4410 [math.CA], 2008-2015.

%H Wolfram Library Archive, KempnerSums.nb (8.6 KB) - Mathematica Notebook, <a href="http://library.wolfram.com/infocenter/MathSource/7166/"> Summing Kempner's Curious (Slowly-Convergent) Series</a>

%F Equals Sum_{j>=1} Sum_{i=0..j-1} 1/(2^i + 2^j).

%F From _Amiram Eldar_, Jun 30 2020: (Start)

%F Equals Sum_{k>=0} 1/(2^k + 1/2).

%F Equals 2 * A323482 - 1. (End)

%F Equals 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

%e Sum_{k>0} 1/A018900(k) = 1.52899956069688841838263949451...

%p evalf( 2*add( (-1)^(n+1)*((4^n + 1)/(4^n - 1))*(1/2)^(n^2), n = 1..18), 100); # _Peter Bala_, Jan 28 2022

%t (* first install irwinSums.m, see either reference, then *) First@ RealDigits@ iSum[1, 2, 2^7, 2]

%Y Cf. A065442, A082830, A082831, A082832, A082833, A082834, A082835, A082836, A082837, A082838, A082839, A140502, A160502, A018900, A323482.

%K base,cons,nonn

%O 1,2

%A _Robert G. Wilson v_, Aug 03 2010