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%I #35 Feb 27 2023 04:01:52
%S 12,17,41,46,70,75,99,104,128,133,157,162,186,191,215,220,244,249,273,
%T 278,302,307,331,336,360,365,389,394,418,423,447,452,476,481,505,510,
%U 534,539,563,568,592,597,621,626,650,655,679,684,708,713,737,742,766
%N Numbers k such that k^2 == -1 (mod 29).
%C Numbers k such that k == 12 or 17 (mod 29). - _Charles R Greathouse IV_, Dec 27 2011
%C 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,...
%H Vincenzo Librandi, <a href="/A155096/b155096.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F From _M. F. Hasler_, Jun 16 2010: (Start)
%F a(n) = 12*(-1)^(n+1) + 29 [n/2].
%F a(2k+1) = 29 k + a(1), a(2k) = 29 k - a(1), with a(1) = A002314(4) since 29 = A002144(4).
%F a(n) = a(n-2) + 29 for all n > 2. (End)
%F G.f.: x*(12 + 5*x + 12*x^2)/((1 + x)*(1 - x)^2). - _Vincenzo Librandi_, May 03 2014
%F Sum_{n>=1} (-1)^(n+1)/a(n) = tan(5*Pi/58)*Pi/29. - _Amiram Eldar_, Feb 27 2023
%e 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.
%t LinearRecurrence[{1,1,-1},{12,17,41},100] (* _Vincenzo Librandi_, Feb 29 2012 *)
%t Select[Range[800], PowerMod[#, 2, 29] == 28 &] (* _Vincenzo Librandi_, Apr 24 2014 *)
%t CoefficientList[Series[(12 + 5 x + 12 x^2)/((1 + x) (1 - x)^2), {x, 0, 30}], x] (* _Vincenzo Librandi_, May 03 2014 *)
%o (PARI) A155096(n)=n\2*29-12*(-1)^n /* _M. F. Hasler_, Jun 16 2010 */
%Y Cf. A002144, A155086, A155095, A155097, A155098.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Jan 20 2009
%E Terms checked & minor edits by _M. F. Hasler_, Jun 16 2010
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