OFFSET
1,1
COMMENTS
The infinite series {s(n)} converges to 1 from below (see link).
LINKS
FORMULA
2^n <= a(n) < 2^(n^2).
EXAMPLE
For n = 3, s(3) = log(16/(sqrt(3)*7^(2/3))) = 0.926009..., so a(3) = floor(1/(1-0.926009...)) = 13.
PROG
(PARI) v = [log(2)]; for(n=2, 32, v = concat(v, v[n-1] + eulerphi(n)/n*log(2^n/(2^n-1)))); a(n) = floor(1/(1-v[n])); \\ can only compute 63 terms due to loss of precision
CROSSREFS
KEYWORD
nonn
AUTHOR
Yifan Xie, Apr 13 2025
STATUS
approved
