

A153438


Least k > 1 such that k^(3^n)*(k^(3^n)+1) + 1 is prime.


13



2, 2, 2, 21, 209, 72, 260, 17, 3311, 4469, 94259, 55599
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OFFSET

0,1


COMMENTS

Numbers of the form k^n*(k^n+1) + 1 with n > 0, k > 1 may be primes only if n has the form 3^j. When n is even k^(4*n) + k^(2*n) + 1 = (k^(2*n)+1)^2  (k^n)^2 = (k^(2*n)+k^n+1)*(k^(2*n)k^n+1) so composite. But why if n odd > 3 and not a power of 3 is k^n*(k^n+1) + 1 always composite?
Phi[3^(n+1),k] = k^(3^n)*(k^(3^n)+1)+1. When m <> 3^n in k^m*(k^m+1)+1, Phi[3m,k] < k^m*(k^m+1)+1 and is a divisor of it.  Lei Zhou, Feb 09 2012
The prime number corresponding to the 10th term is a 587458digit number.  Lei Zhou, Jul 04 2014
x^(2*k) + x^k + 1 = (x^(3*k)  1)/(x^k  1) is the product over n dividing 3k but not dividing k of cyclotomic polynomials Phi(n). If k is a power of 3, n = 3k is the only such divisor and we have a single irreducible cyclotomic polynomial Phi(3k). Otherwise we have the product of more than one polynomial, with integer values > 1 for integer x > 1, and thus always composite numbers.  Martin Becker, Jun 22 2021


LINKS



FORMULA



MATHEMATICA

Table[i = 1; m = 3^u; While[i++; cp = 1 + i^m + i^(2*m); ! PrimeQ[cp]]; i, {u, 1, 7}] (* Lei Zhou, Feb 01 2012 *)


PROG

(PARI) a(n) = my(k=2); while (!isprime(k^(3^n)*(k^(3^n)+1) + 1), k++); k; \\ Michel Marcus, Jan 01 2023


CROSSREFS



KEYWORD

nonn,more,hard


AUTHOR



EXTENSIONS

3311 from Lei Zhou using OpenPFGW, Feb 01 2012
4469 from Lei Zhou using OpenPFGW, Feb 09 2012
New term, 94259, from Lei Zhou using OpenPFGW, Jul 04 2014


STATUS

approved



