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A073668
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Decimal expansion of Sum_{k>=1} 1/(10^k - 1).
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16
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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
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OFFSET
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0,2
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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FORMULA
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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
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EXAMPLE
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0.122324243426244526264428344628264449244... = A065444/9.
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MAPLE
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# 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
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MATHEMATICA
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RealDigits[ N[ Sum[1/(10^k - 1), {k, 1, Infinity}], 120]] [[1]]
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PROG
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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