A253242 Least k>=0 such that n^(2^k)+1 is prime (for even n), or (n^(2^k)+1)/2 is prime (for odd n); -1 if no such k exists.

0, 0, 0, 0, 0, 2, -1, 0, 0, 1, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 2, 1, 0, 1, -1, 0, 1, 0

Offset

2,6

Comments

Least k such that the generalized Fermat number in base n (GFN(k,n)) is prime.

a(n) = -1 if n is in A070265 (perfect powers with an odd exponent).

a(n) is currently unknown for n = {31, 38, 50, 55, 62, 63, 67, 68, 77, 83, 86, 89, 91, 92, 97, 98, 99, 104, 107, 109, 122, 123, 127, 135, 137, ...}

Corresponding primes are {3, 2, 5, 3, 7, 1201, 0, 5, 11, 61, 13, 7, 197, 113, 17, 41761, 19, 181, 401, 11, 23, 139921, 577, 13, 677, 0, 29, 421, 31, ...}. (use 0 if a(n) = -1)

All 2 <= n <= 1500 and 0 <= k <= 14 are checked, the first occurrence of k (start with k = 0) in a(n) are {2, 11, 7, 43, 41, 75, 274, 234, 331, 1342, 824, ...}.

Links

Table of n, a(n) for n=2..30.

Eric Chen, Table of n, a(n) for n = 2..1500 status (for the -1s, only a(n) for n in A070265 are proved, all other -1s are only conjectured)

Gary Barnes, List of generalized Fermat primes in even bases up to 1030

Eric Chen, List of generalized Fermat primes in bases up to 1000

Chris Caldwell, Generalized Fermat number

Richard Fischer, List of generalized Fermat primes in odd bases

Yves Gallot, Generalized Fermat prime search

Wilfrid Keller, Factorization of GFN(n,2)

Wilfrid Keller, Factorization of GFN(n,3)

Wilfrid Keller, Factorization of GFN(n,5)

Wilfrid Keller, Factorization of GFN(n,6)

Wilfrid Keller, Factorization of GFN(n,10)

Wilfrid Keller, Factorization of GFN(n,12)

Jeppe Stig Salling Nielsen, List of generalized Fermat primes in even bases up to 1000

MathWorld, Generalized Fermat number

OEIS wiki, Generalized Fermat number

Wikipedia, Generalized Fermat number

Formula

a(2n) = A228101(n) = log_2(A079706(n)).

a(A006093(n)) = 0, a(A076274(n)) = 0, a(A070265(n)) = -1.

Example

a(7) = 2 since (7^(2^0)+1)/2 and (7^(2^1)+1)/2 are not primes, but (7^(2^2)+1)/2 = 1201 is prime.
a(14) = 1 since 14^(2^0)+1 is not prime, but 14^(2^1)+1 = 197 is prime.

Mathematica

Table[k=0; While[p=If[EvenQ[n], (2n)^(2^k)+1, ((2n)^(2^k)+1)/2]; k<12 && !PrimeQ[p], k=k+1]; If[k==12, -1, k], {n, 2, 1500}]

Prog

(PARI) f(n) = for(k=0, 11, if(ispseudoprime(n^(2^k)+1), return(k))); -1
g(n) = for(k=0, 11, if(ispseudoprime((n^(2^k)+1)/2), return(k))); -1
a(n) = if(n%2==0, f(n), g(n))
(PARI) f(n, k)=if(n%2, (n^(2^k)+1)/2, n^(2^k)+1)
a(n)=if(ispower(-n), -1, my(k); while(!ispseudoprime(f(n, k)), k++); k) \\ Charles R Greathouse IV, Apr 20 2015

Crossrefs

Cf. A079706, A228101, A084712, A123669, A058064, A057856, A130536, A080121, A077659, A122900.

Sequence in context: A350532 A000161 A060398 * A359814 A359815 A260649

Adjacent sequences: A253239 A253240 A253241 * A253243 A253244 A253245

Keyword

sign,more,hard

Author

Eric Chen, Apr 19 2015

Status

approved