|
%I #38 Sep 08 2022 08:46:02
%S 1021,1061,1151,1291,1481,1721,2011,2351,2741,3181,3671,4211,4801,
%T 5441,6131,6871,7661,8501,9391,10331,11321,12361,13451,14591,15781,
%U 17021,18311,19651,21041,22481,23971,25511,27101,28741,30431,32171,33961,35801,37691
%N a(n) = 25*n^2 + 15*n + 1021.
%C This is the case m=5 and k=41 of the formula m^2*n^2 + (m^2 - 2*m)*n + (m^2*k) - (m-1). The most famous example is when m=1 and k=41 (Euler's generating polynomial). With k=41 the formula gives consecutive primes for m=10 and n=0..10, m=17 and n=0..10, m=86 and n=0..8. It is interesting to note that the sequences produced are all factors of the semiprimes produced by m=1, k=41. The other famous values to try for k are 5, 11 and 17 as these all produce primes up to k^2.
%H Vincenzo Librandi, <a href="/A214732/b214732.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: (1021-2002*x+1031*x^2)/(1-x)^3. - _Bruno Berselli_, Aug 28 2012
%F E.g.f.: (1021 + 40*x + 25*x^2)*exp(x). - _G. C. Greubel_, Apr 26 2021
%p A214732:= n-> 25*n^2 +15*n +1021; seq(A214732(n), n=0..40); # _G. C. Greubel_, Apr 26 2021
%t Table[25n^2 +15n +1021, {n, 0, 40}] (* _Vincenzo Librandi_, Aug 29 2012 *)
%o (Magma) [25*n^2+15*n+1021: n in [0..40]] // _Vincenzo Librandi_, Aug 29 2012
%o (PARI) a(n)=25*n^2+15*n+1021 \\ _Charles R Greathouse IV_, Oct 25 2012
%o (Sage) [25*n^2 +15*n +1021 for n in (0..40)] # _G. C. Greubel_, Apr 26 2021
%Y Cf. A215814.
%K nonn,easy
%O 0,1
%A _Robert Potter_, Jul 27 2012
|
