A155095 Numbers k such that k^2 == -1 (mod 17).

4, 13, 21, 30, 38, 47, 55, 64, 72, 81, 89, 98, 106, 115, 123, 132, 140, 149, 157, 166, 174, 183, 191, 200, 208, 217, 225, 234, 242, 251, 259, 268, 276, 285, 293, 302, 310, 319, 327, 336, 344, 353, 361, 370, 378, 387, 395, 404, 412, 421, 429, 438, 446, 455

Offset: 1

Author

Vincenzo Librandi, Jan 20 2009

Keywords

• nonn
• easy

Comments

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

Numbers congruent to {4, 13} mod 17. - Amiram Eldar, Feb 27 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).

Crossrefs

Cf. A002144, A155086, A155096, A155097, A155098.

Programs

Mathematica

Select[Range[500],PowerMod[#,2,17]==16&] (* or *) LinearRecurrence[ {1,1,-1},{4,13,21},60] (* Harvey P. Dale, Jun 25 2011 *)

PARI

A155095(n)=n\2*17-4*(-1)^n /* M. F. Hasler, Jun 16 2010 */

Formula

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

a(n) = 4*(-1)^(n+1) + 17*floor(n/2).

a(2k+1) = 17 k + a(1), a(2k) = 17 k - a(1), with a(1) = A002314(3) since 17 = A002144(3).

a(n) = a(n-2) + 17 for all n > 2. (End)

From Bruno Berselli, Sep 26 2010: (Start)

G.f.: x*(4+9*x+4*x^2)/((1+x)*(1-x)^2).

a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.

a(n) = (34*n + (-1)^n - 17)/4. (End)

Sum_{n>=1} (-1)^(n+1)/a(n) = tan(9*Pi/34)*Pi/17. - Amiram Eldar, Feb 27 2023

Extensions

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