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Primes of the form 4^k + 1 for some k > 0.
4

%I #41 Dec 22 2016 23:20:12

%S 5,17,257,65537

%N Primes of the form 4^k + 1 for some k > 0.

%C Same as Fermat primes 2^(2^m) + 1 for m >= 1. See A019434 for comments, etc.

%C 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.

%C As a(n) is a Fermat prime > 3, by Pépin's test a(n) has primitive root 3.

%C Conjecture: let p a prime number, a(n) is not congruent to p mod (p^2-3)/2. - _Vincenzo Librandi_, Jun 15 2014

%C This conjecture is false when p = a(n), but may be true for primes p != a(n). - _Jonathan Sondow_, Jun 15 2014

%C 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

%D Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966, p. 102, nr. 3.

%D P. L. Chebyshev, Theory of congruences. Elements of number theory, Chelsea, 1972, p. 306.

%H P. L. Chebyshev, <a href="http://archive.org/stream/theoriedercongr00schagoog#page/n330/mode/2up">Theorie der Congruenzen</a>, Mayer & Mueller, 1889, p. 306-313.

%H R. Fueter, <a href="http://dx.doi.org/10.5169/seals-16904">Über primitive Wurzeln von Primzahlen</a>, Comment. Math. Helv., 18 (1946), 217-223, p. 217.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Pépin&#39;s_test">Pépin's test</a>

%H <a href="/index/Ar#Artin">Index entries for sequences related to Artin's conjecture</a>

%H <a href="/index/Pri#primes_root">Index entries for primes by primitive root</a>

%F a(n) = A019434(n+1) for n > 0.

%e 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).

%t Select[Table[4^k + 1, {k, 10^3}], PrimeQ] (* _Michael De Vlieger_, Dec 22 2016 *)

%Y Cf. A005596, A019334, A019434, A090866, A221981.

%K nonn

%O 1,1

%A _Jonathan Sondow_, Feb 04 2013