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%I #51 Jun 09 2015 05:40:49
%S 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,
%T 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,
%U 1,1,2,1,1,1,1,2,1,2,1,1,1,1,2,1,2,1,2
%N 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.
%C a(1) corresponds to the first four Fermat primes. - _Thomas Ordowski_, Apr 22 2012
%C Schinzel's hypothesis H implies that there are arbitrarily large terms in the sequence. - _Thomas Ordowski_, Apr 26 2012
%C First value > 4 is a(102416) = 5 (corresponding to p = 2823521). - _Robert Israel_, May 28 2015
%C First value > 5 is a(4250044) = 6 (corresponding to p = 151062433). - _Giovanni Resta_, Jun 09 2015
%H Charles R Greathouse IV, <a href="/A182199/b182199.txt">Table of n, a(n) for n = 1..10000</a>
%e Let f(p,k) = a^(2^k)+b^(2^k), where f(p,1) = p is a prime of form 4k+1.
%e 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.
%p N:= 10^4: # to get values corresponding to primes <= N
%p Primes:= select(isprime,[4*i+1 $ i=1..floor((N-1)/4)]):
%p G:= map(p -> [Re,Im](GaussInt:-GIfactors(p)[2][1][1]),Primes):
%p f:= proc(ab) local j;
%p for j from 2 do if not isprime(ab[1]^(2^j)+ab[2]^(2^j)) then return(j-1) fi od
%p end proc:
%p map(f,G); # _Robert Israel_, May 28 2015
%t 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 *)
%o (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
%o 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", ")))
%o \\ _Charles R Greathouse IV_, Apr 24 2012
%K nonn
%O 1,1
%A _Thomas Ordowski_, Apr 20 2012
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