A182199 Largest integer N such that a^(2^k) + b^(2^k) for 1 <= k <= N is prime, where p = a^2 + b^2 is the n-th prime of the form 4m+1.

4, 2, 3, 2, 2, 2, 2, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2, 1, 2, 2, 2, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 2

Offset: 1

Author

Thomas Ordowski, Apr 20 2012

Keywords

• nonn

Comments

a(1) corresponds to the first four Fermat primes. - Thomas Ordowski, Apr 22 2012

Schinzel's hypothesis H implies that there are arbitrarily large terms in the sequence. - Thomas Ordowski, Apr 26 2012

First value > 4 is a(102416) = 5 (corresponding to p = 2823521). - Robert Israel, May 28 2015

First value > 5 is a(4250044) = 6 (corresponding to p = 151062433). - Giovanni Resta, Jun 09 2015

Examples

			Let f(p,k) = a^(2^k)+b^(2^k), where f(p,1) = p is a prime of form 4k+1.
f(5,1) = 5, f(5,2) = 17, f(5,3) = 257, f(5,4) = 65537, f(5,5) = 641*6700417. So N = 4. Next prime of form 4k+1 is 13; N = 2. 17; N = 3. etc.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Programs

Maple

N:= 10^4: # to get values corresponding to primes <= N
Primes:= select(isprime,[4*i+1 $ i=1..floor((N-1)/4)]):
G:= map(p -> [Re,Im](GaussInt:-GIfactors(p)[2][1][1]),Primes):
f:= proc(ab) local j;
  for j from 2 do if not isprime(ab[1]^(2^j)+ab[2]^(2^j)) then return(j-1) fi od
end proc:
map(f,G); # Robert Israel, May 28 2015

Mathematica

nn = 35; pr = {}; Do[p = a^2 + b^2; If[p < nn^2 && PrimeQ[p], AppendTo[pr, {p, a, b}]], {a, nn}, {b, a}]; pr = Sort[pr]; {jnk, a, b} = Transpose[pr]; Table[i = 1; While[PrimeQ[a[[n]]^2^i + b[[n]]^2^i], i++]; i - 1, {n, 2, Length[pr]}] (* T. D. Noe, Apr 24 2012 *)

PARI

f(p)=my(s=lift(sqrt(Mod(-1, p))), x=p, t); if(s>p/2, s=p-s); while(s^2>p, t=s; s=x%s; x=t); s
forprime(p=5,1e3,if(p%4==1,a=f(p);b=sqrtint(p-a^2);n=1; while(ispseudoprime(a^(2^n)+b^(2^n)),n++);print1(n-1", ")))
\\ Charles R Greathouse IV, Apr 24 2012