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 A153438 Least k > 1 such that k^(3^n)*(k^(3^n)+1) + 1 is prime. 13
 2, 2, 21, 209, 72, 260, 17, 3311, 4469, 94259 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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 587458-digit 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 Lei Zhou, Prime Database Entry, July 4th, 2014. 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 *) CROSSREFS Cf. A101406, A153436, A056993. Sequence in context: A081687 A082811 A014353 * A190632 A036110 A143807 Adjacent sequences:  A153435 A153436 A153437 * A153439 A153440 A153441 KEYWORD nonn,more,hard AUTHOR Pierre CAMI, Dec 26 2008 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 Name and Comment corrected by Robert Price, Nov 11 2018 STATUS approved

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Last modified January 24 13:11 EST 2022. Contains 350538 sequences. (Running on oeis4.)