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A153438
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Least k > 1 such that k^(3^n)*(k^(3^n)+1) + 1 is prime.
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13
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2, 2, 2, 21, 209, 72, 260, 17, 3311, 4469, 94259, 55599
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OFFSET
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0,1
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COMMENTS
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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
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LINKS
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FORMULA
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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n) = my(k=2); while (!isprime(k^(3^n)*(k^(3^n)+1) + 1), k++); k; \\ Michel Marcus, Jan 01 2023
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CROSSREFS
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KEYWORD
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nonn,more,hard
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AUTHOR
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EXTENSIONS
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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
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STATUS
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approved
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