OFFSET
0,2
COMMENTS
Parallels A000005 up to a(46).
Sum_{k>=1} x^k/(1-x^k) = Sum_{k>=1} tau(k)*x^k. Choosing x = 1/10 gives the result. - Amarnath Murthy, Oct 21 2002
REFERENCES
Amarnath Murthy, Some interesting results on d(N), the number of divisors of a natural number, page 463, Octogon Mathematical Magazine, Vol. 8 No. 2, October 2000.
FORMULA
From Eric Desbiaux, Mar 11 2009: (Start)
Equals Sum_{k >= 1} 1/((2^k*5^k)-1).
Equals Sum_{k >= 1} (1/2^k)*(1/5^k)/(1-((1/2^k)*(1/5^k))).
Sum_{k >= 1} 1/(5^k) = 1/4.
Sum_{k >= 1} 1/(2^k) = 1.
Sum_{k >= 1} (1/5^k)/(1-((1/2^k)*(1/5^k))) = 0.2726344339156...
Sum_{k >= 1} (1/2^k)/(1-((1/2^k)*(1/5^k))) = 1.0582125127815...
Sum_{k >= 1} 1/(1-((1/2^k)*(1/5^k))) - 1 = A073668.
(End)
Fast computation via Lambert series: 0.122324243426... = Sum_{n>=1} x^(n^2)*(1+x^n)/(1-x^n) where x=1/10. - Joerg Arndt, Oct 18 2020
EXAMPLE
0.122324243426244526264428344628264449244... = A065444/9.
MAPLE
evalf(Sum(1/(10^k - 1), k = 1..infinity), 200) # Vaclav Kotesovec, Jul 16 2019
# second program with faster converging series after Joerg Arndt
evalf( add( (1/10)^(n^2)*(1 + 2/(10^n - 1)), n = 1..8), 105); # Peter Bala, Jan 30 2022
MATHEMATICA
RealDigits[ N[ Sum[1/(10^k - 1), {k, 1, Infinity}], 120]] [[1]]
PROG
(PARI) suminf(k=1, 1/(10^k-1)) \\ Charles R Greathouse IV, Oct 05 2014
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
Robert G. Wilson v, Aug 29 2002
STATUS
approved