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A158803 Numbers k such that k^2 == 2 (mod 41). 6
17, 24, 58, 65, 99, 106, 140, 147, 181, 188, 222, 229, 263, 270, 304, 311, 345, 352, 386, 393, 427, 434, 468, 475, 509, 516, 550, 557, 591, 598, 632, 639, 673, 680, 714, 721, 755, 762, 796, 803, 837, 844, 878, 885, 919, 926, 960, 967, 1001, 1008, 1042, 1049 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Numbers congruent to {17, 24} mod 41. - Amiram Eldar, Feb 26 2023
LINKS
FORMULA
a(n) = a(n-1) + a(n-2) - a(n-3).
a(n) = (1/4)*(41 + 27*(-1)^(n-1) + 82*(n-1)).
First differences: a(2n) - a(2n-1) = 7, a(2n+1) - a(2n) = 34.
G.f.: x*(17 + 7*x + 17*x^2)/((1+x)*(x-1)^2). - R. J. Mathar, Apr 04 2009
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(7*Pi/82)*Pi/41. - Amiram Eldar, Feb 26 2023
MATHEMATICA
LinearRecurrence[{1, 1, -1}, {17, 24, 58}, 60] (* Vincenzo Librandi, Mar 02 2012 *)
Select[Range[1200], PowerMod[#, 2, 41]==2&] (* Harvey P. Dale, Oct 24 2021 *)
PROG
(Magma) I:=[17, 24, 58]; [n le 3 select I[n] else Self(n-1)+Self(n-2)-Self(n-3): n in [1..60]]; // Vincenzo Librandi, Mar 02 2012
(PARI) a(n) = (1/4)*(41+27*(-1)^(n-1)+82*(n-1)); \\ Vincenzo Librandi, Mar 02 2012
CROSSREFS
Sequence in context: A070687 A205697 A231963 * A340050 A051780 A124971
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 27 2009
EXTENSIONS
Comments translated to formulas by R. J. Mathar, Apr 04 2009
STATUS
approved

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Last modified August 19 05:14 EDT 2024. Contains 375284 sequences. (Running on oeis4.)