cp's OEIS Frontend

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

%I A155095 #32 Feb 27 2023 04:01:36
%S A155095 4,13,21,30,38,47,55,64,72,81,89,98,106,115,123,132,140,149,157,166,
%T A155095 174,183,191,200,208,217,225,234,242,251,259,268,276,285,293,302,310,
%U A155095 319,327,336,344,353,361,370,378,387,395,404,412,421,429,438,446,455
%N A155095 Numbers k such that k^2 == -1 (mod 17).
%C A155095 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...
%C A155095 Numbers congruent to {4, 13} mod 17. - _Amiram Eldar_, Feb 27 2023
%H A155095 Vincenzo Librandi, <a href="/A155095/b155095.txt">Table of n, a(n) for n = 1..1000</a>
%H A155095 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F A155095 From _M. F. Hasler_, Jun 16 2010: (Start)
%F A155095 a(n) = 4*(-1)^(n+1) + 17*floor(n/2).
%F A155095 a(2k+1) = 17 k + a(1), a(2k) = 17 k - a(1), with a(1) = A002314(3) since 17 = A002144(3).
%F A155095 a(n) = a(n-2) + 17 for all n > 2. (End)
%F A155095 From _Bruno Berselli_, Sep 26 2010: (Start)
%F A155095 G.f.: x*(4+9*x+4*x^2)/((1+x)*(1-x)^2).
%F A155095 a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
%F A155095 a(n) = (34*n + (-1)^n - 17)/4. (End)
%F A155095 Sum_{n>=1} (-1)^(n+1)/a(n) = tan(9*Pi/34)*Pi/17. - _Amiram Eldar_, Feb 27 2023
%t A155095 Select[Range[500],PowerMod[#,2,17]==16&] (* or *) LinearRecurrence[ {1,1,-1},{4,13,21},60] (* _Harvey P. Dale_, Jun 25 2011 *)
%o A155095 (PARI) A155095(n)=n\2*17-4*(-1)^n /* _M. F. Hasler_, Jun 16 2010 */
%Y A155095 Cf. A002144, A155086, A155096, A155097, A155098.
%K A155095 nonn,easy
%O A155095 1,1
%A A155095 _Vincenzo Librandi_, Jan 20 2009
%E A155095 Terms checked & minor edits by _M. F. Hasler_, Jun 16 2010