A090106 - OEIS

%I #19 Sep 27 2024 07:22:41

%S 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,

%T 27,219

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

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

%C a(29) > 5*10^9, if it exists. - _Amiram Eldar_, Sep 27 2024

%e 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.

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

%o (PARI) isp(x) = isprime(x^2 + x + 41);

%o 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

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

%K nonn,more

%O 1,2

%A _Labos Elemer_, Dec 22 2003

%E 2 wrong terms removed by _Amiram Eldar_, Sep 27 2024