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a(n) = (1/4)*n^4 - (1/2)*n^3 + (3/4)*n^2 - (1/2)*n + 41.
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%I #28 Sep 08 2022 08:46:13

%S 41,43,53,83,151,281,503,853,1373,2111,3121,4463,6203,8413,11171,

%T 14561,18673,23603,29453,36331,44351,53633,64303,76493,90341,105991,

%U 123593,143303,165283,189701,216731,246553,279353,315323,354661,397571,444263,494953

%N a(n) = (1/4)*n^4 - (1/2)*n^3 + (3/4)*n^2 - (1/2)*n + 41.

%C Empirical Observation: Reasonably productive (better than 85% in first 24 terms) prime-generating polynomial.

%C All integers generated by this polynomial for 0 < n <= 24 are prime with the exception of a(14) = 47*179, a(17) = 71*263, and a(20) = 47*773.

%C Negative and zero values of n also produce primes but they are not unique.

%C a(n) = (1/4)*n^4 - (1/2)*n^3 + (3/4)*n^2 - (1/2)*n + 809, for 0 < n <= 24, is also reasonably productive but produces composites at a(4), a(7), a(19) and a(20).

%C a(n) = (1/4)*n^4 - (1/2)*n^3 + (3/4)*n^2 - (1/2)*n + 641, for 0 < n <= 24, is also quite productive.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html">Prime-Generating Polynomial</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: x*(41 - 162*x + 248*x^2 - 162*x^3 + 41*x^4)/(1-x)^5. - _Vincenzo Librandi_, Jul 03 2015

%F a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5), n>5. - _Wesley Ivan Hurt_, Jul 09 2015

%p A259552:=n->n^4/4-n^3/2+3*n^2/4-n/2+41: seq(A259552(n), n=1..100); # _Wesley Ivan Hurt_, Jul 09 2015

%t f[n_] := n^4/4 - n^3/2 + 3 n^2/4 - n/2 + 41; Array[f, 38] (* or *)

%t LinearRecurrence[{5, -10, 10, -5, 1}, {41, 43, 53, 83, 151}, 38] (* _Robert G. Wilson v_, Jul 07 2015 *)

%o (Magma) [(1/4)*n^4-(1/2)*n^3+(3/4)*n^2-(1/2)*n+41: n in [1..40]]; // _Vincenzo Librandi_, Jul 03 2015

%Y Cf. A202018.

%K nonn,easy

%O 1,1

%A _Robert Potter_, Jun 30 2015

%E Corrected and extended by _Vincenzo Librandi_, Jul 03 2015