%I #35 Dec 14 2023 06:08:42
%S 1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,
%T 1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,
%U 1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6,1,6
%N Repeat (1,6): Period 2.
%C Continued fraction for (3+sqrt(15))/6. - _Philippe Deléham_, Sep 25 2006
%C This sequence can be generated by an infinite number of formulas all having the form a^(b*n) mod c subject to the following conditions. The number a can be congruent to either 3,5 or 6 mod 7. If a is congruent to 3 or 5 mod 7 then b can be any number of the form 3*k+6. If a is congruent to 6 mod 7 then b can be any number of the form 2k+1. Finally, if a is congruent to either 6, 26, or 31 mod 35 then c can be 7 or 35; otherwise, we use c = 7. For example: a(n) = 33^(15*n) mod 7, a(n) = 31^(9*n) mod 7, and a(n) = 31^(9*n) mod 35. - _Gary Detlefs_, May 19 2014
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,1).
%F G.f.: (1+6*x)/(1-x^2). - _Philippe Deléham_, Sep 25 2006
%F a(n) = 6^n mod 7. - _Zerinvary Lajos_, Nov 26 2009
%p A010687:=n->(6^n mod 7); seq(A010687(n), n=0..100); # _Wesley Ivan Hurt_, May 19 2014
%t Table[Mod[6^n, 7], {n, 0, 100}] (* _Wesley Ivan Hurt_, May 19 2014 *)
%o (Sage) [power_mod(6,n,7)for n in range(0,100)] # _Zerinvary Lajos_, Nov 26 2009
%o (PARI) a(n)=n%2*5+1 \\ _Charles R Greathouse IV_, Jul 13 2016
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Dec 11 1996