OFFSET
0,1
COMMENTS
For n >= 2, the Fermat number F(n) = 2^(2^n) + 1 is prime if and only if F(n) divides a(2^n-2).
REFERENCES
Kusta Inkeri, Tests for primality, Ann. Acad. Sci. Fenn., A I No. 279 (1960), pp. 1-19.
M. Krizek, F. Luca, L. Somer, 17 Lectures on Fermat Numbers: From Number Theory to Geometry, CMS Books in Mathematics, vol. 9, Springer-Verlag, New York, 2001, p. 46.
FORMULA
For n >= 2, a(n) = 2 + Sum_{k=1..2^(n-1)} (-1)^k*64^k*2^(n-1)*binomial(k + 2^(n-1) - 1, 2*k - 1)/k.
a(n) = ceiling(c^(2^n)) where c = 4 + sqrt(15) is the largest root of x^2 - 8*x + 1 = 0.
a(n) = (4 + sqrt(15))^(2^n) + (4 - sqrt(15))^(2^n).
a(n) = 2*T(2^n,4), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
sqrt(5/3) = Product_{n >= 0} (1 + 2/a(n)).
2*sqrt(5/3)/3 = Product_{n >= 0} (1 - 1/a(n)).
a(n) = 2*A005828(n).
MATHEMATICA
NestList[#^2 - 2 &, 8, 6]
PROG
(Magma) [8] cat [n eq 1 select 62 else Self(n-1)^2-2: n in [1..6]];
(PARI) {a(n)=if(n<1, 8*(n==0), a(n-1)^2-2)};
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Arkadiusz Wesolowski, Jul 26 2021
STATUS
approved