|
| |
|
|
A155095
|
|
Numbers k such that k^2 == -1 (mod 17).
|
|
7
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
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...
|
|
|
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
|
a(n) = 4*(-1)^(n+1) + 17*floor(n/2). - M. F. Hasler, Jun 16 2010
a(2k+1) = 17 k + a(1), a(2k) = 17 k - a(1), with a(1) = A002314(3) since 17 = A002144(3). - M. F. Hasler, Jun 16 2010
a(n) = a(n-2) + 17 for all n > 2. - M. F. Hasler, Jun 16 2010
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)
|
|
|
MATHEMATICA
|
Select[Range[500], PowerMod[#, 2, 17]==16&] (* or *) LinearRecurrence[ {1, 1, -1}, {4, 13, 21}, 60] (* Harvey P. Dale, Jun 25 2011 *)
|
|
|
PROG
|
(PARI) A155095(n)=n\2*17-4*(-1)^n /* M. F. Hasler, Jun 16 2010 */
|
|
|
CROSSREFS
|
Cf. A002144, A155086, A155096, A155097, A155098.
Sequence in context: A339271 A081024 A339216 * A063219 A063121 A199798
Adjacent sequences: A155092 A155093 A155094 * A155096 A155097 A155098
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
Vincenzo Librandi, Jan 20 2009
|
|
|
EXTENSIONS
|
Terms checked & minor edits by M. F. Hasler, Jun 16 2010
|
|
|
STATUS
|
approved
|
| |
|
|
