A155096 Numbers k such that k^2 == -1 (mod 29).

12, 17, 41, 46, 70, 75, 99, 104, 128, 133, 157, 162, 186, 191, 215, 220, 244, 249, 273, 278, 302, 307, 331, 336, 360, 365, 389, 394, 418, 423, 447, 452, 476, 481, 505, 510, 534, 539, 563, 568, 592, 597, 621, 626, 650, 655, 679, 684, 708, 713, 737, 742, 766

Offset: 1

Author

Vincenzo Librandi, Jan 20 2009

Keywords

• nonn
• easy

Comments

Numbers k such that k == 12 or 17 (mod 29). - Charles R Greathouse IV, Dec 27 2011

The first pair (a,b) is such that a+b=p, a*b=p*h+1, with h<=(p-1)/4; subsequent pairs are given as (a+kp, b+kp), k=1,2,3,...

Examples

			Let p = 29, a+b=29, a*b=29h+1, h<=7; for h=7, a+b=29, a*b=204, a=12, b=17; other pairs (12+29, 17+29) and so on.
		

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. A002144, A155086, A155095, A155097, A155098.

Programs

Mathematica

LinearRecurrence[{1,1,-1},{12,17,41},100] (* Vincenzo Librandi, Feb 29 2012 *)
Select[Range[800], PowerMod[#, 2, 29] == 28 &] (* Vincenzo Librandi, Apr 24 2014 *)
CoefficientList[Series[(12 + 5 x + 12 x^2)/((1 + x) (1 - x)^2), {x, 0, 30}], x] (* Vincenzo Librandi, May 03 2014 *)

PARI

A155096(n)=n\2*29-12*(-1)^n /* M. F. Hasler, Jun 16 2010 */

Formula

From M. F. Hasler, Jun 16 2010: (Start)

a(n) = 12*(-1)^(n+1) + 29 [n/2].

a(2k+1) = 29 k + a(1), a(2k) = 29 k - a(1), with a(1) = A002314(4) since 29 = A002144(4).

a(n) = a(n-2) + 29 for all n > 2. (End)

G.f.: x*(12 + 5*x + 12*x^2)/((1 + x)*(1 - x)^2). - Vincenzo Librandi, May 03 2014

Sum_{n>=1} (-1)^(n+1)/a(n) = tan(5*Pi/58)*Pi/29. - Amiram Eldar, Feb 27 2023

Extensions

Terms checked & minor edits by M. F. Hasler, Jun 16 2010