0,1

a(n) is the ratio of two consecutive base 3 Fermat numbers A129290(n) = 3^(3^n) + 1 = {4, 28, 19684, 7625597484988, ...}.

a(n) = A002061(3^(3^n)). a(n) = A129290(n+1) / A129290(n).

Table[1 - 3^3^n + 9^3^n, {n, 0, 5}]

Cf. A129290 = 3^(3^n) + 1. Cf. A055777 = 3^(3^n). Cf. A002061 = Central polygonal numbers: n^2 - n + 1.

Sequence in context: A038803 A144957 A163016 this_sequence A159815 A070746 A068004

Adjacent sequences: A129288 A129289 A129290 this_sequence A129292 A129293 A129294

nonn

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