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A129290 Base-3 Fermat numbers: 3^(3^n) + 1. 4
4, 28, 19684, 7625597484988, 443426488243037769948249630619149892804, 87189642485960958202911070585860771696964072404731750085525219437990967093723439943475549906831683116791055225665628 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
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).
FORMULA
a(n) = 3^(3^n) + 1. a(n) = A055777(n) + 1.
MATHEMATICA
Table[3^3^n+1, {n, 0, 6}]
PROG
(PARI) a(n)=3^(3^n)+1 \\ Charles R Greathouse IV, Jul 05 2011
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).
Sequence in context: A218174 A220756 A202713 * A307553 A327436 A339266
KEYWORD
nonn
AUTHOR
Alexander Adamchuk, Apr 08 2007
STATUS
approved

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Last modified April 23 08:33 EDT 2024. Contains 371905 sequences. (Running on oeis4.)