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A383024
a(n) = floor(1/(1-s(n))) where s(n) = Sum_{k=1..n} phi(k)/k * log(2^k/(2^k-1)). phi(k) is the Euler totient function A000010(k).
1
3, 6, 13, 23, 61, 90, 229, 417, 917, 1429, 3916, 5748, 16323, 28489, 53121, 89322, 249079, 364536, 1068096, 1802530, 3542367, 5749605, 16694858, 24980973, 61771834, 107398135, 230197552, 363961698, 1053601058, 1426995284, 3997869809, 7478431118
OFFSET
1,1
COMMENTS
The infinite series {s(n)} converges to 1 from below (see link).
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
Cf. A000010.
Sequence in context: A323580 A002799 A285263 * A162426 A374627 A058554
KEYWORD
nonn
AUTHOR
Yifan Xie, Apr 13 2025
STATUS
approved