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A155096
Numbers k such that k^2 == -1 (mod 29).
6
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,...
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
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Jan 20 2009
EXTENSIONS
Terms checked & minor edits by M. F. Hasler, Jun 16 2010
STATUS
approved