%I #36 Feb 28 2023 02:13:29
%S 8,23,39,54,70,85,101,116,132,147,163,178,194,209,225,240,256,271,287,
%T 302,318,333,349,364,380,395,411,426,442,457,473,488,504,519,535,550,
%U 566,581,597,612,628,643,659,674,690,705,721,736,752,767,783,798,814
%N Numbers k such that k^2 == 2 (mod 31).
%C Sequences of the type n^2 == 2 (mod m) are basically defined for each m of A057126. See A047341 (m=7), A113804 (m=14), A155449 (m=17), A155450 (m=23), A158803 (m=41) etc. - _R. J. Mathar_, Aug 26 2009
%H Vincenzo Librandi, <a href="/A164131/b164131.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F a(n) = a(n-1)+a(n-2)-a(n-3).
%F a(n) = (31+(-1)^(n-1)+62(n-1))/4.
%F G.f.: x*(8+15*x+8*x^2)/((1+x)*(x-1)^2). - _R. J. Mathar_, Aug 26 2009
%F a(n) = 31*(n-1)-a(n-1) with n>1, a(1)=8. - _Vincenzo Librandi_, Nov 30 2010
%F Sum_{n>=1} (-1)^(n+1)/a(n) = tan(15*Pi/62)*Pi/31. - _Amiram Eldar_, Feb 28 2023
%e At n= 4, a(4)=(31-1+186)/4=54. At n=5, a(5)=(31+1+248)/4=70.
%t Select[Range[850],Mod[#^2,31]==2&] (* _Harvey P. Dale_, Feb 04 2011 *)
%o (PARI) isok(k) = Mod(k, 31)^2 == 2; \\ _Michel Marcus_, Nov 22 2022
%Y Cf. A057126, A047341, A113804, A155449, A155450, A158803.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Aug 11 2009
%E Entries checked by _R. J. Mathar_, Aug 26 2009