A073668 Decimal expansion of Sum_{k>=1} 1/(10^k - 1).

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, 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, 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

Offset: 0

Author

Robert G. Wilson v, Aug 29 2002

Keywords

• cons
• nonn

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

Examples

			0.122324243426244526264428344628264449244... = A065444/9.
		

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.

Crossrefs

Cf. A065442, A214369, A248721, A248722, A248723, A248724, A248725, A248726.

Programs

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]]

PARI

suminf(k=1,1/(10^k-1)) \\ Charles R Greathouse IV, Oct 05 2014

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