_Paul D. Hanna (pauldhanna(AT)juno.com), _, Aug 22 2006
_Paul D. Hanna (pauldhanna(AT)juno.com), _, Aug 22 2006
Consists entirely of 3's, 5's, and 7's, after an initial partial quotient of 4. These partial quotients are aperiodic.
cofr,nonn,new
cofr,nonn,new
Paul D . Hanna (pauldhanna(AT)juno.com), Aug 22 2006
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
0,2
Consists entirely of 3's, 5's, and 7's, after an initial partial quotient of 4. These partial quotients are aperiodic.
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 (A122164) 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), ...
(PARI) {a(n)=local(x=sum(k=0, ceil(3+log(n+1)), 1/5^(2^k))); contfrac(x)[n+1]}
Cf. A122164.
cofr,nonn
Paul D Hanna (pauldhanna(AT)juno.com), Aug 22 2006
approved