A158803 Numbers k such that k^2 == 2 (mod 41).

17, 24, 58, 65, 99, 106, 140, 147, 181, 188, 222, 229, 263, 270, 304, 311, 345, 352, 386, 393, 427, 434, 468, 475, 509, 516, 550, 557, 591, 598, 632, 639, 673, 680, 714, 721, 755, 762, 796, 803, 837, 844, 878, 885, 919, 926, 960, 967, 1001, 1008, 1042, 1049

Offset

1,1

Comments

Numbers congruent to {17, 24} mod 41. - Amiram Eldar, Feb 26 2023

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Formula

a(n) = a(n-1) + a(n-2) - a(n-3).

a(n) = (1/4)*(41 + 27*(-1)^(n-1) + 82*(n-1)).

First differences: a(2n) - a(2n-1) = 7, a(2n+1) - a(2n) = 34.

G.f.: x*(17 + 7*x + 17*x^2)/((1+x)*(x-1)^2). - R. J. Mathar, Apr 04 2009

Sum_{n>=1} (-1)^(n+1)/a(n) = tan(7*Pi/82)*Pi/41. - Amiram Eldar, Feb 26 2023

Mathematica

LinearRecurrence[{1, 1, -1}, {17, 24, 58}, 60] (* Vincenzo Librandi, Mar 02 2012 *)
Select[Range[1200], PowerMod[#, 2, 41]==2&] (* Harvey P. Dale, Oct 24 2021 *)

Prog

(Magma) I:=[17, 24, 58]; [n le 3 select I[n] else Self(n-1)+Self(n-2)-Self(n-3): n in [1..60]]; // Vincenzo Librandi, Mar 02 2012
(PARI) a(n) = (1/4)*(41+27*(-1)^(n-1)+82*(n-1)); \\ Vincenzo Librandi, Mar 02 2012

Crossrefs

Sequence in context: A070687 A205697 A231963 * A340050 A051780 A124971

Adjacent sequences: A158800 A158801 A158802 * A158804 A158805 A158806

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 27 2009

Extensions

Comments translated to formulas by R. J. Mathar, Apr 04 2009

Status

approved