cp's OEIS Frontend

A157116 Numbers k such that k^2 + 1 == 0 (mod 41^2).

%I A157116 #30 Feb 26 2023 02:33:08
%S A157116 378,1303,2059,2984,3740,4665,5421,6346,7102,8027,8783,9708,10464,
%T A157116 11389,12145,13070,13826,14751,15507,16432,17188,18113,18869,19794,
%U A157116 20550,21475,22231,23156,23912,24837,25593,26518,27274,28199,28955,29880
%N A157116 Numbers k such that k^2 + 1 == 0 (mod 41^2).
%H A157116 Vincenzo Librandi, <a href="/A157116/b157116.txt">Table of n, a(n) for n = 1..1000</a>
%H A157116 Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H A157116 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F A157116 a(1)=378, a(2)=1303; a(n) = 2*a(n-1) - a(n-2) - 13^2 if n is odd, and a(n) = 2*a(n-1) - a(n-2) + 13^2 if n is even.
%F A157116 From _R. J. Mathar_, Mar 08 2009: (Start)
%F A157116 a(n) = (3362n - 1681 + 169*(-1)^n)/4.
%F A157116 G.f.: x*(14*x+27)*(27*x+14)/((1+x)*(x-1)^2). (End)
%F A157116 Sum_{n>=1} (-1)^(n+1)/a(n) = cot(378*Pi/1681)*Pi/1681. - _Amiram Eldar_, Feb 26 2023
%e A157116 378^2 + 1 == 0 (mod 41^2).
%e A157116 1303^2 + 1 == 0 (mod 41^2).
%e A157116 2059^2 + 1 == 0 (mod 41^2).
%t A157116 CoefficientList[Series[(14 x + 27) (27 x + 14) / ((1 + x) (x - 1)^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Sep 11 2013 *)
%t A157116 Select[Range[30000],PowerMod[#,2,1681]==1680&] (* or *) LinearRecurrence[ {1,1,-1},{378,1303,2059},40] (* _Harvey P. Dale_, Jul 05 2021 *)
%o A157116 (Magma) [(3362*n-1681+169*(-1)^n)/4: n in [1..40]]; // _Vincenzo Librandi_, Sep 11 2013
%Y A157116 Cf. A157105, A157111.
%K A157116 nonn,easy
%O A157116 1,1
%A A157116 _Vincenzo Librandi_, Feb 23 2009