OFFSET
1,1
COMMENTS
Dual constant: A119812 = Sum_{n>=1} A049472(n)/2^n = Sum_{n>=1} 1/2^A001951(n). Since this constant c = 2 + Sum_{n>=1} 1/2^A003151(n), where A003151(n) = n + floor(n*sqrt(2)), then the binary expansion of the fractional part of c has 1's only at positions given by Beatty sequence A003151(n) and zeros elsewhere. Plouffe's Inverter describes approximations to the fractional part of c as "polylogarithms type of series with the floor function [ ]."
LINKS
Kevin O'Bryant, A generating function technique for Beatty sequences and other step sequences, Journal of Number Theory, Volume 94, Issue 2, June 2002, Pages 299-319.
FORMULA
Equals Sum(1/(2^q-1)) with the summation extending over all pairs of integers gcd(p,q) = 1, 0 < p/q < sqrt(2) (O'Bryant, 2002). - Amiram Eldar, May 25 2023
EXAMPLE
c = 2.32258852258806773012144068278798408011950250800432925665718...
Continued fraction (A119810):
c = [2;3,10,132,131104,2199023259648,633825300114114700748888473600,..]
where partial quotients are given by:
The following are equivalent expressions for the constant:
where [x] = floor(x).
These series illustrate the above expressions:
(1) c = 1/2^0 + 1/2^1 + 1/2^2 + 1/2^2 + 1/2^3 + 1/2^4 + 1/2^4 +...
(2) c = 1/2^1 + 2/2^2 + 4/2^3 + 5/2^4 + 7/2^5 + 8/2^6 + 9/2^7 +...
(3) c = 2 + 1/2^2 + 1/2^4 + 1/2^7 + 1/2^9 + 1/2^12 + 1/2^14 +...
(4) c =-2 + 1/2^0 + 1/2^0 + 1/2^1 + 1/2^1 + 1/2^2 + 1/2^2 + 1/2^2 +...
(5) c = 1 + 1/2^1 + 2/2^2 + 1/2^3 + 2/2^4 + 1/2^5 + 1/2^6 + 2/2^7 +...
PROG
(PARI) {a(n)=local(t=sqrt(2), x=sum(m=1, 10*n, floor(m*t)/2^m)); floor(10^n*x)%10}
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
Paul D. Hanna, May 26 2006
STATUS
approved