A129290 Base-3 Fermat numbers: 3^(3^n) + 1.

4, 28, 19684, 7625597484988, 443426488243037769948249630619149892804, 87189642485960958202911070585860771696964072404731750085525219437990967093723439943475549906831683116791055225665628

Offset: 0

Author

Alexander Adamchuk, Apr 08 2007

Keywords

• nonn

Comments

Apparently discovered (with arbitrary base) by Gottschalk in 1938 and independently by Ferentinou-Nicolacopoulou in 1963. - Charles R Greathouse IV, Jul 05 2011

a(n) divides a(n+1). a(n+1)/a(n) = (3^(3^(n+1)) + 1)/(3^(3^n) + 1) = 1 - 3^(3^n) + 9^(3^n) = A002061(3^(3^n)) = A129291(n) = {7, 703, 387400807, 58149737003032434092905183, ...}.

References

• J. Ferentinou-Nicolacopoulou, "Une propriété des diviseurs du nombre r^(r^m)+1. Applications au dernier théorème de Fermat." Bulletin Société Mathématique de Grèce 4:1 (1963), pp. 121-126.

Links

• Eugen Gottschalk, Zum Fermatschen Problem, Mathematische Annalen 115 (1934), pp. 157-158.
• Lorenzo Sauras-Altuzarra, Some properties of the factors of Fermat numbers, Art Discrete Appl. Math. (2022).

Crossrefs

Cf. A000215 (Fermat numbers: 2^(2^n) + 1).

Cf. A055777 (3^(3^n)).

Cf. A129291 (A129290(n+1) / A129290(n)).

Cf. A002061 (central polygonal numbers: n^2 - n + 1).

Programs

Mathematica

Table[3^3^n+1,{n,0,6}]

PARI

a(n)=3^(3^n)+1 \\ Charles R Greathouse IV, Jul 05 2011

Formula

a(n) = 3^(3^n) + 1. a(n) = A055777(n) + 1.