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%I #49 Oct 29 2023 01:51:14
%S 1,2,2,3,2,4,2,4,3,4,2,6,2,4,4,5,2,6,2,6,4,4,2,8,3,4,4,6,2,8,2,6,4,4,
%T 4,9,2,4,4,8,2,8,2,6,6,4,3,0,3,6,4,6,2,8,4,8,4,4,3,2,2,4,6,7,4,8,2,6,
%U 4,8,3,2,2,4,6,6,4,8,3,0,5,4,3,2,4,4,4,8,3,2,4,6,4,4,5,2,2,6,6,9,2,8,2,8,8
%N Decimal expansion of Sum_{k>=1} 1/(10^k - 1).
%C Parallels A000005 up to a(46).
%C 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
%D 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.
%F From _Eric Desbiaux_, Mar 11 2009: (Start)
%F Equals Sum_{k >= 1} 1/((2^k*5^k)-1).
%F Equals Sum_{k >= 1} (1/2^k)*(1/5^k)/(1-((1/2^k)*(1/5^k))).
%F Sum_{k >= 1} 1/(5^k) = 1/4.
%F Sum_{k >= 1} 1/(2^k) = 1.
%F Sum_{k >= 1} (1/5^k)/(1-((1/2^k)*(1/5^k))) = 0.2726344339156...
%F Sum_{k >= 1} (1/2^k)/(1-((1/2^k)*(1/5^k))) = 1.0582125127815...
%F Sum_{k >= 1} 1/(1-((1/2^k)*(1/5^k))) - 1 = A073668.
%F (End)
%F 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
%e 0.122324243426244526264428344628264449244... = A065444/9.
%p evalf(Sum(1/(10^k - 1), k = 1..infinity), 200) # _Vaclav Kotesovec_, Jul 16 2019
%p # second program with faster converging series after _Joerg Arndt_
%p evalf( add( (1/10)^(n^2)*(1 + 2/(10^n - 1)), n = 1..8), 105); # _Peter Bala_, Jan 30 2022
%t RealDigits[ N[ Sum[1/(10^k - 1), {k, 1, Infinity}], 120]] [[1]]
%o (PARI) suminf(k=1,1/(10^k-1)) \\ _Charles R Greathouse IV_, Oct 05 2014
%Y Cf. A065442, A214369, A248721, A248722, A248723, A248724, A248725, A248726.
%K cons,nonn
%O 0,2
%A _Robert G. Wilson v_, Aug 29 2002
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