0,2
2^n+1 divides 2^(2^n)-1 iff n is a power of 2.
a(n) = (2^(2^(2^n))-1)/(2^(2^n)+1). a(n) = A051179(2^n)/A000215(n).
Table[ (2^2^2^n - 1) / (2^2^n + 1), {n, 0, 3} ]
Cf. A000215 = Fermat numbers: 2^(2^n)+1. Cf. A051179 = 2^(2^n)-1.
Sequence in context: A200735 A196628 A089895 * A136544 A024048 A094319
Adjacent sequences: A116210 A116211 A116212 * A116214 A116215 A116216
nonn
Alexander Adamchuk (alex(AT)kolmogorov.com), Apr 08 2007
