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

1,1

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,...

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

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

Example

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.

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 *)

Prog

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

Crossrefs

Cf. A002144, A155086, A155095, A155097, A155098.

Sequence in context: A214514 A188004 A045699 * A166674 A243609 A082241

Adjacent sequences: A155093 A155094 A155095 * A155097 A155098 A155099

Keyword

nonn,easy

Author

Vincenzo Librandi, Jan 20 2009

Extensions

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

Status

approved