A164131 Numbers k such that k^2 == 2 (mod 31).

8, 23, 39, 54, 70, 85, 101, 116, 132, 147, 163, 178, 194, 209, 225, 240, 256, 271, 287, 302, 318, 333, 349, 364, 380, 395, 411, 426, 442, 457, 473, 488, 504, 519, 535, 550, 566, 581, 597, 612, 628, 643, 659, 674, 690, 705, 721, 736, 752, 767, 783, 798, 814

Offset: 1

Author

Vincenzo Librandi, Aug 11 2009

Keywords

• nonn
• easy

Comments

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

Examples

			At n= 4, a(4)=(31-1+186)/4=54. At n=5, a(5)=(31+1+248)/4=70.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A057126, A047341, A113804, A155449, A155450, A158803.

Programs

Mathematica

Select[Range[850],Mod[#^2,31]==2&]  (* Harvey P. Dale, Feb 04 2011 *)

PARI

isok(k) = Mod(k, 31)^2 == 2; \\ Michel Marcus, Nov 22 2022

Formula

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

a(n) = (31+(-1)^(n-1)+62(n-1))/4.

G.f.: x*(8+15*x+8*x^2)/((1+x)*(x-1)^2). - R. J. Mathar, Aug 26 2009

a(n) = 31*(n-1)-a(n-1) with n>1, a(1)=8. - Vincenzo Librandi, Nov 30 2010

Sum_{n>=1} (-1)^(n+1)/a(n) = tan(15*Pi/62)*Pi/31. - Amiram Eldar, Feb 28 2023

Extensions

Entries checked by R. J. Mathar, Aug 26 2009