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%I #61 Feb 21 2024 10:52:57
%S 2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,
%T 2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,2,4,
%U 2,4,2,4,2,4,2,4,2,4,2,4,2
%N Period 2: repeat (2,4).
%C Simple continued fraction expansion of 1 + sqrt(3/2) = A176051. - _R. J. Mathar_, Mar 08 2012
%C Number of linear characters of dihedral group of order 2(n+1). - _Eric M. Schmidt_, Feb 12 2013
%C a(n) is the n-th increment between two consecutive elements of the wheel in the wheel factorization with the basis {2, 3}. See A038179. - _Wojciech Raszka_, May 10 2019
%C In base 3, make a sequence such that after the initial term 2, each term is the sum of the squares of the digits of the previous term. That's this sequence (see A000216 for the base 10 version). - _Alonso del Arte_, Mar 19 2020
%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=1073">Encyclopedia of Combinatorial Structures 1073</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,1).
%F From _R. J. Mathar_, Aug 28 2008: (Start)
%F a(n) = 2 * A000034(n).
%F G.f.: 2(1 + 2x)/((1 - x)(1 + x)). (End)
%F a(n) = a(n-2) for n >= 2. - _Jaume Oliver Lafont_, Mar 20 2009
%F a(n) = 2^(n+1) mod 6. - _Roderick MacPhee_, Mar 31 2011
%t ContinuedFraction[1 + Sqrt[6]/2, 100] (* _Alonso del Arte_, Mar 25 2020 *)
%o (PARI) a(n)=2*(1+n%2) \\ _Jaume Oliver Lafont_, Mar 20 2009
%o (Scala) (0 to 99).map(_ % 2 * 2 + 2) // _Alonso del Arte_, Mar 25 2020
%Y Cf. A000034, A038179, A176051.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_
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