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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. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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

Schinzel's hypothesis H says that can be arbitrarily large terms in the sequence. - Thomas Ordowski, Apr 26 2012

LINKS

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

EXAMPLE

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.

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 *)

PROG

(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

CROSSREFS

Sequence in context: A018845 A028947 A068152 * A217435 A238352 A079636

Adjacent sequences:  A182196 A182197 A182198 * A182200 A182201 A182202

KEYWORD

nonn

AUTHOR

Thomas Ordowski, Apr 20 2012

STATUS

approved

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Last modified November 27 15:29 EST 2014. Contains 250224 sequences.