OFFSET
0,1
COMMENTS
The polynomial generates 31 primes in row starting from n=0.
The polynomial 16*n^2 - 668*n + 7013 generates the same primes in reverse order.
Note: all the polynomials of the form p^2*n^2 +- p*n + 41, p^2*n^2 +- 3*p*n + 43, p^2*n^2 +- 5*p*n + 47, ..., p^2*n^2 +- (2k+1)*p*n + q, ..., p^2*n^2 +- 79*p*n + 1601, where q is a (prime) term of the Euler polynomial q = k^2 + k + 41, from k=0 to k=39, have their discriminant equal to -163*p^2; the demonstration is easy: the discriminant is equal to b^2 - 4ac = (2k+1)^2*p^2 - 4*q*p^2 = - p^2 ((2k+1)^2 - 4q) = - p^2*(4k^2 + 4k + 1 - 4k^2 - 4k - 164) = -163*p^2.
Observation: many of the polynomials formed this way have the capacity to generate many primes in row. Examples:
9n^2 + 3n + 41 generates 27 primes in row starting from n=0 (and 40 primes for n = n-13);
9n^2 - 237n + 1601 generates 27 primes in row starting from n=0;
16n^2 + 4n + 41 generates, for n = n-21 (that is 16n^2 - 668n + 7013) 31 primes in row.
LINKS
Bruno Berselli, Table of n, a(n) for n = 0..1000
M. Coman, Ten prime-generating quadratic polynomials, Preprint 2015.
Factor Database, Factorizations of 16n^2-292n+1373.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
G.f.: (1373-3022*x+1681*x^2)/(1-x)^3. - Bruno Berselli, Apr 06 2012
MATHEMATICA
Table[16*n^2 - 292*n + 1373, {n, 0, 50}] (* T. D. Noe, Apr 04 2012 *)
PROG
(Magma) [n^2-73*n+1373: n in [0..172 by 4]]; // Bruno Berselli, Apr 06 2012
(PARI) a(n)=16*n^2-292*n+1373 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Marius Coman, Apr 04 2012
EXTENSIONS
Offset changed from 1 to 0 by Bruno Berselli, Apr 06 2012
STATUS
approved