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Prime-generating polynomial: 4n^2 + 12n - 1583.
1

%I #17 Sep 08 2022 08:45:55

%S -1583,-1567,-1543,-1511,-1471,-1423,-1367,-1303,-1231,-1151,-1063,

%T -967,-863,-751,-631,-503,-367,-223,-71,89,257,433,617,809,1009,1217,

%U 1433,1657,1889,2129,2377,2633,2897,3169,3449,3737,4033,4337,4649,4969,5297,5633,5977

%N Prime-generating polynomial: 4n^2 + 12n - 1583.

%C The polynomial generates 35 primes/negative values of primes in row starting from n=0.

%C The polynomial 4*n^2 - 284*n + 3449 generates the same primes in reverse order.

%C Other related polynomials:

%C For n = 6n+6 than n = n-11 we get 144n^2 - 2808n + 12097 which generates 16 primes in row starting from n=0 (with the discriminant equal to 2^9*3^2*199);

%C For n = 12n+12 than n = n-15 we get 576n^2 - 15984n + 109297 which generates 17 primes in row starting from n=0 (with the discriminant equal to 2^11*3^2*199).

%C So this polynomial opens at least two directions of study:

%C (1) polynomials of type 4n^2 + 12n - p, where p is prime (could be of the form 30k+23);

%C (2) polynomials with the discriminant equal to 2^n*3^m*199, where n is odd and m is even (an example of such a polynomial, with the discriminant equal to 2^5*3^4*199, is 36n^2 - 1020n + 3643 which generates 32 primes for values of n from 0 to 34).

%H Vincenzo Librandi, <a href="/A182409/b182409.txt">Table of n, a(n) for n = 0..1000</a>

%H E. W. Weisstein, <a href="http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html">MathWorld: Prime-Generating Polynomial</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F From _Chai Wah Wu_, May 28 2016: (Start)

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F G.f.: (1591*x^2 - 3182*x + 1583)/(x - 1)^3.

%F (End)

%t Table[4 n^2 + 12 n - 1583, {n, 0, 50}] (* _Vincenzo Librandi_, May 29 2016 *)

%o (PARI) a(n)=4*n^2+12*n-1583 \\ Charles R Greathouse IV, Oct 01 2012

%o (Magma) [4*n^2+12*n-1583: n in [0..50]]; // _Vincenzo Librandi_, May 29 2016

%K sign,easy

%O 0,1

%A _Marius Coman_, May 09 2012