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.

Adjacent sequences: A116210 A116211 A116212 this_sequence A116214 A116215 A116216

Sequence in context: A134909 A003831 A089895 this_sequence A136544 A024048 A094319

nonn

Alexander Adamchuk (alex(AT)kolmogorov.com), Apr 08 2007