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A014113 a(n) = a(n-1) + 2*a(n-2) with a(0)=0, a(1)=2.

%I A014113 #80 May 18 2025 19:41:24
%S A014113 0,2,2,6,10,22,42,86,170,342,682,1366,2730,5462,10922,21846,43690,
%T A014113 87382,174762,349526,699050,1398102,2796202,5592406,11184810,22369622,
%U A014113 44739242,89478486,178956970,357913942,715827882,1431655766,2863311530,5726623062,11453246122
%N A014113 a(n) = a(n-1) + 2*a(n-2) with a(0)=0, a(1)=2.
%H A014113 Vincenzo Librandi, <a href="/A014113/b014113.txt">Table of n, a(n) for n = 0..1000</a>
%H A014113 INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=457">Encyclopedia of Combinatorial Structures 457</a>
%H A014113 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,2).
%F A014113 a(0) = 0 and if n>=1, a(n) = 2^n - a(n-1).
%F A014113 a(n) = A078008(n+1). - _Reinhard Zumkeller_, Jun 10 2005
%F A014113 a(n) = 2*A001045(n). - _Benoit Jubin_, Dec 01 2011
%F A014113 a(n) = (2^(n+1) - 2*(-1)^n)/3. - _Ralf Stephan_, Jul 18 2013
%F A014113 G.f.: 2*x/(1+x)/(1-2*x). - _Colin Barker_, Feb 19 2012
%F A014113 G.f.: 1/Q(0) -1, where Q(k) = 1 + 2*x^2 - (2*k+3)*x + x*(2*k+1 - 2*x)/Q(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Oct 05 2013
%F A014113 Consider a Pascal-like triangle T(n,k) in which T(n,0) = T(n,n) = 0 if n even, 1 if n odd, and each interior entry T(n,k) is the sum of the two entries "above" it: T(n,k) = T(n-1,k-1) + T(n-1,k) for 0 < k < n. Then, a(n) is the sum of the entries in the n-th row of T(n,k). - _Greg Dresden_, May 24 2024
%t A014113 LinearRecurrence[{1,2},{0,2},50] (* _Vincenzo Librandi_, Feb 19 2012 *)
%o A014113 (Haskell)
%o A014113 a014113 n = a014113_list !! n
%o A014113 a014113_list = 0 : 2 : zipWith (+)
%o A014113                        (map (* 2) a014113_list) (tail a014113_list)
%o A014113 -- _Reinhard Zumkeller_, Jul 20 2013
%Y A014113 Cf. A001045, A078008, A163271, A290732.
%K A014113 nonn,easy
%O A014113 0,2
%A A014113 _N. J. A. Sloane_