A078886 Decimal expansion of Sum {n>=0} 1/5^(2^n).
2, 4, 1, 6, 0, 2, 5, 6, 0, 0, 0, 6, 5, 5, 3, 6, 0, 0, 0, 0, 0, 0, 4, 2, 9, 4, 9, 6, 7, 2, 9, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 8, 4, 4, 6, 7, 4, 4, 0, 7, 3, 7, 0, 9, 5, 5, 1, 6, 1, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 4, 0, 2, 8, 2, 3, 6, 6, 9, 2, 0, 9, 3, 8, 4
Offset: 0
Examples
0.241602560006553600000... From _Paul D. Hanna_, Aug 22 2006: (Start) Decimal expansion consists of large gaps of zeros between strings of digits that form powers of 2; this can be seen by grouping the digits as follows: x = .2 4 16 0 256 000 65536 000000 4294967296 000000000000 ...= 0.24160256000655360000004294... and then recognizing the substrings as powers of 2: 2 = 2^(2^0), 4 = 2^(2^1), 16 = 2^(2^2), 65536 = 2^(2^4), 4294967296 = 2^(2^5), 18446744073709551616 = 2^(2^6), ... (End)
Links
- Aubrey J. Kempner, On Transcendental Numbers, Transactions of the American Mathematical Society, volume 17, number 4, October 1916, pages 476-482.
- Index entries for transcendental numbers
Crossrefs
Programs
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Mathematica
RealDigits[ N[ Sum[1/5^(2^n), {n, 0, Infinity}], 110]][[1]]
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PARI
{a(n)=local(x=sum(k=0,ceil(3+log(n+1)),1/5^(2^k)));(floor(10^n*x))%10} \\ Paul D. Hanna, Aug 22 2006
Formula
Equals -Sum_{k>=1} mu(2*k)/(5^k - 1), where mu is the Möbius function (A008683). - Amiram Eldar, Jul 12 2020
Extensions
Edited by R. J. Mathar, Aug 02 2008
Comments