A122165 Continued fraction expansion of constant x = Sum_{n>=0} 1/5^(2^n).
0, 4, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 3, 5, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 5, 3, 7, 5, 3, 5, 5, 7, 5, 3, 7, 5, 5, 3, 5, 7, 3, 5, 7, 5, 3, 5, 5, 7, 3, 5, 7, 5, 5, 3, 5, 7, 5
Offset: 0
Examples
x=[0;4,7,5,5,3,5,7,5,3,7,5,3,5,5,7,5,3,7,5,5,3,5,7,3,5,7,5,3,5,5,7,5,...]. x=0.2416025600065536000000429496729600000000000018446744073709551616000... Decimal expansion (A078886) 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 ... 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), ...
Crossrefs
Cf. A078886.
Programs
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PARI
{a(n)=local(x=sum(k=0,ceil(3+log(n+1)),1/5^(2^k)));contfrac(x)[n+1]}
Comments