OFFSET
1,1
COMMENTS
Same as Fermat primes 2^(2^m) + 1 for m >= 1. See A019434 for comments, etc.
Chebyshev showed that 3 is a primitive root of all primes of the form 2^(2*k) + 1 with k > 0. If the sequence is infinite, then Artin's conjecture ("every nonsquare integer n != -1 is a primitive root of infinitely many primes q") is true for n = 3.
As a(n) is a Fermat prime > 3, by Pépin's test a(n) has primitive root 3.
Conjecture: let p a prime number, a(n) is not congruent to p mod (p^2-3)/2. - Vincenzo Librandi, Jun 15 2014
This conjecture is false when p = a(n), but may be true for primes p != a(n). - Jonathan Sondow, Jun 15 2014
Primes p with the property that k-th smallest divisor of its squares p^2, for all 1 <= k <= tau(p^2), contains exactly k "1" digits in its binary representation. Corresponding values of squares p^2: 25, 289, 66049, 4295098369. Example: p = 257, set of divisors of p^2 = 66049 in binary representation: {1, 100000001, 10000001000000001}. Subsequence of A255401. - Jaroslav Krizek, Dec 21 2016
REFERENCES
Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966, p. 102, nr. 3.
P. L. Chebyshev, Theory of congruences. Elements of number theory, Chelsea, 1972, p. 306.
LINKS
P. L. Chebyshev, Theorie der Congruenzen, Mayer & Mueller, 1889, p. 306-313.
R. Fueter, Über primitive Wurzeln von Primzahlen, Comment. Math. Helv., 18 (1946), 217-223, p. 217.
Wikipedia, Pépin's test
FORMULA
a(n) = A019434(n+1) for n > 0.
EXAMPLE
4^1 + 1 = 5 is prime, so a(1) = 5. Also, 3^k == 3, 4, 2, 1 (mod 5) for k = 1, 2, 3, 4, resp., so 3 is a primitive root for a(1).
MATHEMATICA
Select[Table[4^k + 1, {k, 10^3}], PrimeQ] (* Michael De Vlieger, Dec 22 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Feb 04 2013
STATUS
approved