OFFSET
1,1
COMMENTS
From Peter Bala, Apr 15 2018: (Start)
The polynomial P(n) := n^2 + n + 11 takes distinct prime values for the 10 consecutive integers n = 0 to 9. It follows that the polynomial P(n-10) = (n - 10)^2 + (n - 10) + 11 takes prime values for the 20 consecutive integers n = 0 to 19, consisting of the 10 primes above each taken twice. We note two consequences of this fact.
1) The polynomial P(2*n-10) = 4*n^2 - 38*n + 101 also takes prime values for the 10 consecutive integers n = 0 to 9.
2)The polynomial P(3*n-10) = 9*n^2 - 57*n + 101 takes prime values for the 7 consecutive integers n = 0 to 6 (= floor(19/3)). In addition, calculation shows that P(3*n-10) also takes prime values for n from -3 to -1. Equivalently put, the polynomial P(3*n-19) = 9*n^2 - 111*n + 353 takes prime values for the 10 consecutive integers n = 0 to 9. Cf. A007635 and A005846. (End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Prime-Generating Polynomial
MATHEMATICA
lst={}; Do[p=n^2+n+11; If[PrimeQ[p], AppendTo[lst, p]], {n, 0, 5*5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Jan 27 2009 *)
Select[Table[n^2+n+11, {n, 0, 600}], PrimeQ] (* Vincenzo Librandi, Dec 07 2011 *)
PROG
(Magma) [ a: n in [0..200] | IsPrime(a) where a is n^2+n+11 ]; // Vincenzo Librandi, Dec 07 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved