A248721 Decimal expansion of Sum_{k>=1} 1/(4^k - 1).
4, 2, 1, 0, 9, 7, 6, 8, 6, 0, 3, 3, 4, 2, 3, 7, 7, 7, 2, 9, 5, 9, 9, 0, 8, 8, 7, 9, 6, 7, 7, 1, 3, 0, 4, 8, 9, 6, 1, 4, 4, 1, 3, 3, 6, 3, 2, 4, 1, 1, 5, 4, 0, 4, 6, 0, 5, 9, 2, 0, 7, 9, 6, 7, 1, 2, 7, 7, 1, 3, 7, 0, 4, 8, 8, 7, 3, 9, 8, 0, 2, 7, 5, 1, 9, 0, 3, 6, 8, 4, 7, 5, 8, 6, 5, 0, 7, 9, 5, 3, 9, 2, 8, 4, 5
Offset: 0
Examples
0.4210976860334237772959908879677130489614413363241154046059207967127713704887...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..10000
Programs
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Maple
evalf(sum(1/(4^k-1), k=1..infinity),120) # Vaclav Kotesovec, Oct 18 2014 # second program with faster converging series after Joerg Arndt evalf( add( (1/4)^(n^2)*(1 + 2/(4^n - 1)), n = 1..13), 105); # Peter Bala, Jan 30 2022
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Mathematica
x = 1/4; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* after an observation and the formula of Amarnath Murthy, see A073668 *) -
PARI
suminf(k=1, 1/(4^k-1)) \\ Michel Marcus, Oct 18 2014
Formula
Equals Sum_{k>=1} x^(k^2)*(1+x^k)/(1-x^k) where x = 1/4 (the Lambert series evaluated at 1/4). - Joerg Arndt, Jun 03 2020
Equals Sum_{k>=1} d(k)/4^k, where d(k) is the number of divisors of k (A000005). - Amiram Eldar, Jun 22 2020