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%I #65 Dec 14 2023 06:08:27
%S 1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,
%T 1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,1,5,
%U 1,5,1,5,1,5,1,5,1,5,1,5,1
%N Periodic sequence: repeat [1, 5].
%C Also continued fraction expansion of (5+3*sqrt(5))/10. - _Bruno Berselli_, Sep 30 2011
%C From _Gary Detlefs_, May 19 2014: (Start)
%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 is congruent to either 5,11,13,17,21, or 23 mod 24 and b is of the form 2k+1.
%C 1. If a = 5 mod 6 then c = 6.
%C 2. If a = 5 mod 8 then c = 8.
%C 3. If a = 5 mod 12 then c = 12.
%C 4. If a = 5 mod 24 then c = 24.
%C For example: a(n)= 13^(5*n) mod 8, a(n)= 29^(7*n) mod c where c is any number in {6,8,12,24}. (End)
%H Burkard Polster, <a href="http://plus.maths.org/issue52/features/polster/index.html">Juggling, maths and a beautiful mind</a> [From Parthasarathy Nambi, Nov 20 2009]
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,1).
%F From _Paul Barry_, Jun 03 2003: (Start)
%F G.f.: (1+5*x)/((1-x)*(1+x)).
%F E.g.f.: 3*exp(x)-2*exp(-x).
%F a(n) = 3-2(-1)^n.
%F a(n) = 5^((1-(-1)^n)/2) = 5^(ceiling(n/2)-floor(n/2)). (End)
%F a(n) = 5^n mod 24. - _Paul Curtz_, Jan 09 2008
%F a(n) = 5^n mod 12. - _Zerinvary Lajos_, Nov 25 2009
%F a(n) = A000364(n+1) mod 10. - _Paul Curtz_, Feb 09 2010
%F a(n) = 11^n mod 6. - _Vincenzo Librandi_, Jun 01 2016
%p [seq (modp((4*n+1),8),n=0..80)]; # _Zerinvary Lajos_, Dec 01 2006
%t PadRight[{},120,{1,5}] (* _Harvey P. Dale_, Aug 19 2012 *)
%o (Sage) [pow(5,n,12) for n in range(51)] # _Zerinvary Lajos_, Nov 25 2009
%o (Maxima) A010686(n):=if evenp(n) then 1 else 5$
%o makelist(A010686(n),n,0,30); /* _Martin Ettl_, Nov 09 2012 */
%o (PARI) a(n)=n%2*4+1 \\ _Charles R Greathouse IV_, Sep 24 2015
%Y Cf. A000364.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
%E Definition rewritten by _Bruno Berselli_, Sep 30 2011
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