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%I #64 Sep 08 2022 08:45:01
%S 1,5,14,37,97,254,665,1741,4558,11933,31241,81790,214129,560597,
%T 1467662,3842389,10059505,26336126,68948873,180510493,472582606,
%U 1237237325,3239129369,8480150782,22201322977,58123818149,152170131470,398386576261,1042989597313
%N Expansion of (1+2*x)/(1-3*x+x^2).
%C Binomial transform of A000285. - _R. J. Mathar_, Oct 26 2011
%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.
%H G. C. Greubel, <a href="/A054486/b054486.txt">Table of n, a(n) for n = 0..1000</a>
%H I. Adler, <a href="http://www.fq.math.ca/Scanned/7-2/adler.pdf">Three Diophantine equations - Part II</a>, Fib. Quart., 7 (1969), pp. 181-193.
%H E. I. Emerson, <a href="http://www.fq.math.ca/Scanned/7-3/emerson.pdf">Recurrent Sequences in the Equation DQ^2=R^2+N</a>, Fib. Quart., 7 (1969), pp. 231-242.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,-1).
%F a(n) = 3*a(n-1) - a(n-2), a(0)=1, a(1)=5.
%F a(n) = (5*(((3+sqrt(5))/2)^n - ((3-sqrt(5))/2)^n) - (((3+sqrt(5))/2)^(n-1) - ((3-sqrt(5))/2)^(n-1)))/sqrt(5).
%F a(n) + 7*A001519(n) = A005248(n). - _Creighton Dement_, Oct 30 2004
%F a(n) = Lucas(2*n+1) + Fibonacci(2*n) = A002878(n) + A001906(n) = A025169(n-1) + A001906(n+1).
%F a(n) = (-1)^n*Sum_{k = 0..n} A238731(n,k)*(-6)^k. - _Philippe Deléham_, Mar 05 2014
%F 0 = -11 + a(n)^2 - 3*a(n)*a(n+1) + a(n+1)^2 for all n in Z. - _Michael Somos_, Mar 17 2015
%F a(n) = -2*F(n)^2 + 6*F(n)*F(n+1) + F(n+1)^2 for all n in Z where F = Fibonacci. - _Michael Somos_, Mar 17 2015
%F a(n) = 3*F(2*n) + F(2*n+1) for all n in Z where F = Fibonacci. - _Michael Somos_, Mar 17 2015
%F a(n) = -A100545(-2-n) for all n in Z. - _Michael Somos_, Mar 17 2015
%F a(n) = A000285(2*n) = A228208(2*n+1) = A104449(2*n+1) for all n in Z. - _Michael Somos_, Mar 17 2015
%F From _Klaus Purath_, Nov 05 2019: (Start)
%F a(n) = (a(n-m) + a(n+m))/Lucas(2*m), m <= n.
%F a(n) = sum of 2*m+1 consecutive terms starting with a(n-m) divided by Lucas(2*m+1), m <= n.
%F a(n) = alternating sum of 2*m+1 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+1), m <= n.
%F a(n) + a(n+1) = sum of 2*m+2 consecutive terms starting with a(n-m) divided by Fibonacci(2*m+2), m <= n.
%F a(n) + a(n+1) = (a(n-m) + a(n+m+1))/Fibonacci(2*m+1), m <= n.
%F The following formulas are extended to negative indexes:
%F a(n) = 3*Fibonacci(2*n+1) - Fibonacci(2*n-3).
%F a(n) = (Fibonacci(2*n+5) - 3* Fibonacci(2*n-1))/2.
%F a(n) = (4*Lucas(2*n+2) - Lucas(2*n-4))/5.
%F a(n) = Fibonacci(2*n+5) - 4*Fibonacci(2*n+1).
%F a(n) = (5*Fibonacci(2*n+5) - Fibonacci(2*n-7))/12. (End)
%F E.g.f.: exp(-(1/2)*(-3+sqrt(5))*x)*(-7 + sqrt(5) + (7 + sqrt(5))*exp(sqrt(5)*x))/(2*sqrt(5)). - _Stefano Spezia_, Nov 19 2019
%F a(n) = 3*n + 1 + Sum_{k=1..n} k*a(n-k). - _Yu Xiao_, Jun 20 2020
%e G.f. = 1 + 5*x + 14*x^2 + 37*x^3 + 97*x^4 + 254*x^5 + 665*x^6 + 1741*x^7 + ...
%p with(combinat); f:=fibonacci; seq(f(2*n+2)+2*f(2*n), n=0..30); # _G. C. Greubel_, Nov 08 2019
%t CoefficientList[Series[(2*z+1)/(z^2-3*z+1), {z, 0, 30}], z] (* _Vladimir Joseph Stephan Orlovsky_, Jul 15 2011 *)
%t a[ n_]:= 3 Fibonacci[2n] + Fibonacci[2n+1]; (* _Michael Somos_, Mar 17 2015 *)
%t LinearRecurrence[{3,-1},{1,5},40] (* _Harvey P. Dale_, Apr 24 2019 *)
%o (PARI) Vec((1+2*x)/(1-3*x+x^2)+O(x^99)) \\ _Charles R Greathouse IV_, Jul 15 2011
%o (PARI) {a(n) = 3*fibonacci(2*n) + fibonacci(2*n+1)}; /* _Michael Somos_, Mar 17 2015 */
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1+2*x)/(1-3*x+x^2)) ); // _Marius A. Burtea_, Nov 05 2019
%o (Magma) a:=[1,5]; [n le 2 select a[n] else 3*Self(n-1)-Self(n-2): n in [1..30]]; // _Marius A. Burtea_, Nov 05 2019
%o (Sage) f=fibonacci; [f(2*n+2) +2*f(2*n) for n in (0..30)] # _G. C. Greubel_, Nov 08 2019
%o (GAP) F:=Fibonacci;; List([0..30], n-> F(2*n+2) +2*F(2*n) ); # _G. C. Greubel_, Nov 08 2019
%Y Cf. A000032, A000045, A000285, A002878, A100545, A104449, A228208.
%K easy,nonn
%O 0,2
%A _Barry E. Williams_, May 06 2000
%E "a(1)=5", not "a(0)=5" from Dan Nielsen (nielsed(AT)uah.edu), Sep 10 2009
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