A181969 - OEIS

A181969

Prime-generating polynomial: 16*n^2 - 292*n + 1373.

1373, 1097, 853, 641, 461, 313, 197, 113, 61, 41, 53, 97, 173, 281, 421, 593, 797, 1033, 1301, 1601, 1933, 2297, 2693, 3121, 3581, 4073, 4597, 5153, 5741, 6361, 7013, 7697, 8413, 9161, 9941, 10753, 11597, 12473, 13381, 14321, 15293, 16297, 17333, 18401

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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