A007404 Decimal expansion of Sum_{n>=0} 1/2^(2^n).
8, 1, 6, 4, 2, 1, 5, 0, 9, 0, 2, 1, 8, 9, 3, 1, 4, 3, 7, 0, 8, 0, 7, 9, 7, 3, 7, 5, 3, 0, 5, 2, 5, 2, 2, 1, 7, 0, 3, 3, 1, 1, 3, 7, 5, 9, 2, 0, 5, 5, 2, 8, 0, 4, 3, 4, 1, 2, 1, 0, 9, 0, 3, 8, 4, 3, 0, 5, 5, 6, 1, 4, 1, 9, 4, 5, 5, 5, 3, 0, 0, 0, 6, 0, 4, 8, 5, 3, 1, 3, 2, 4, 8, 3, 9, 7, 2, 6, 5, 6, 1, 7, 5, 5, 8
Offset: 0
Examples
0.81642150902189314370....
References
- M. J. Knight, An "oceans of zeros" proof that a certain non-Liouville number is transcendental, The American Mathematical Monthly, Vol. 98, No. 10 (1991), pp. 947-949.
Links
- Harry J. Smith, Table of n, a(n) for n = 0..20000
- Boris Adamczewski, The Many Faces of the Kempner Number, Journal of Integer Sequences, Vol. 16 (2013), #13.2.15.
- David H. Bailey, Jonathan M. Borwein, Richard E. Crandall, and Carl Pomerance, On the Binary Expansions of Algebraic Numbers, Journal de Théorie des Nombres de Bordeaux, volume 16, number 3, 2004, pages 487-518. Also LBNL-53854 2003, and authors' copies one, four.
- D. H. Bailey and H. R. P. Ferguson, Numerical results on relations between fundamental constants using a new algorithm, Mathematics of Computation, Vol.53 No. 188 (1989), 649-656. (Annotated scanned copy)
- F. R. Bernhart & N. J. A. Sloane, Emails, April-May 1994
- Aubrey J. Kempner, On transcendental numbers, Transactions of the American Mathematical Society 17 (1916), pp. 476-482.
- Simon Plouffe, Plouffe's Inverter, sum(1/2^(2^n), n=0..infinity); to 20000 digits
- Simon Plouffe, sum(1/2^(2^n), n=0..infinity) to 1024 digits
- Jeffrey Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217.
- Index entries for transcendental numbers
Programs
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Mathematica
RealDigits[ N[ Sum[1/2^(2^n), {n, 0, Infinity}], 110]] [[1]]
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PARI
default(realprecision, 20080); x=suminf(n=0, 1/2^(2^n)); x*=10; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b007404.txt", n, " ", d)); \\ Harry J. Smith, May 07 2009
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PARI
suminf(k = 0, 1/(2^(2^k))) \\ Michel Marcus, Mar 26 2017
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PARI
suminf(k=0,1.>>2^k) \\ Charles R Greathouse IV, Nov 07 2017
Formula
Equals -Sum_{k>=1} mu(2*k)/(2^k - 1) = Sum_{k>=1, k odd} mu(k)/(2^k - 1). - Amiram Eldar, Jun 22 2020
Extensions
Edited by Robert G. Wilson v, Dec 11 2002
Deleted old PARI program Harry J. Smith, May 20 2009
Comments