A090106 Values of k such that {P(k), P(k+1), ..., P(k+12)} are all prime numbers, where P(k) = k^2 + k + 41.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 219

Offset

1,2

Comments

a(n) is the first argument providing 13 "polynomially consecutive" primes with respect to the polynomial x^2 + x + 41.

a(29) > 5*10^9, if it exists. - Amiram Eldar, Sep 27 2024

Links

Table of n, a(n) for n=1..28.

Example

k = 219: {P(219), ..., P(231)} = {48221, ..., 53633}, i.e., 13 consecutive integer values substituted to P(x) = x^2 + x + 41 polynomial, all provide primes. The "classical case" includes one single 41-chain of PC-primes, see A055561.

Mathematica

Position[Times @@@ Partition[Table[Boole@PrimeQ[k^2 + k + 41], {k, 1, 1000}], 13, 1], 1] // Flatten (* Amiram Eldar, Sep 27 2024 *)

Prog

(PARI) isp(x) = isprime(x^2 + x + 41);
lista(kmax) = {my(v = vector(13, k, isp(k))); for(k = 14, kmax, if(vecprod(v) == 1, print1(k - 13, ", ")); v = concat(vecextract(v, "^1"), isp(k))); } \\ Amiram Eldar, Sep 27 2024

Crossrefs

Cf. A055561, A090101, A090102, A090562, A090563.

Sequence in context: A154470 A194907 A389006 * A369464 A167662 A309810

Adjacent sequences: A090103 A090104 A090105 * A090107 A090108 A090109

Keyword

nonn,more

Author

Labos Elemer, Dec 22 2003

Extensions

2 wrong terms removed by Amiram Eldar, Sep 27 2024

Status

approved