A179951 Decimal expansion of Sum_{k has exactly two bits equal to 1 in base 2} 1/k.

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, 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, 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

Offset: 1

Author

Robert G. Wilson v, Aug 03 2010

Keywords

• base
• cons
• nonn

Comments

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.

Examples

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

Links

• Robert Baillie, Summing The Curious Series Of Kempner and Irwin, arXiv:0806.4410 [math.CA], 2008-2015.
• Wolfram Library Archive, KempnerSums.nb (8.6 KB) - Mathematica Notebook, Summing Kempner's Curious (Slowly-Convergent) Series

Crossrefs

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

Programs

Maple

 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

Mathematica

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

Formula

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

From Amiram Eldar, Jun 30 2020: (Start)

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

Equals 2 * A323482 - 1. (End)

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