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

378, 1303, 2059, 2984, 3740, 4665, 5421, 6346, 7102, 8027, 8783, 9708, 10464, 11389, 12145, 13070, 13826, 14751, 15507, 16432, 17188, 18113, 18869, 19794, 20550, 21475, 22231, 23156, 23912, 24837, 25593, 26518, 27274, 28199, 28955, 29880

Offset: 1

Author

Vincenzo Librandi, Feb 23 2009

Keywords

• nonn
• easy

Examples

			378^2 + 1 == 0 (mod 41^2).
1303^2 + 1 == 0 (mod 41^2).
2059^2 + 1 == 0 (mod 41^2).
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A157105, A157111.

Programs

Magma

[(3362*n-1681+169*(-1)^n)/4: n in [1..40]]; // Vincenzo Librandi, Sep 11 2013

Mathematica

CoefficientList[Series[(14 x + 27) (27 x + 14) / ((1 + x) (x - 1)^2), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 11 2013 *)
Select[Range[30000],PowerMod[#,2,1681]==1680&] (* or *) LinearRecurrence[ {1,1,-1},{378,1303,2059},40] (* Harvey P. Dale, Jul 05 2021 *)

Formula

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.

From R. J. Mathar, Mar 08 2009: (Start)

a(n) = (3362n - 1681 + 169*(-1)^n)/4.

G.f.: x*(14*x+27)*(27*x+14)/((1+x)*(x-1)^2). (End)

Sum_{n>=1} (-1)^(n+1)/a(n) = cot(378*Pi/1681)*Pi/1681. - Amiram Eldar, Feb 26 2023