%I #71 Jul 29 2024 06:23:19
%S 1,4,20,104,544,2848,14912,78080,408832,2140672,11208704,58689536,
%T 307302400,1609056256,8425127936,44114542592,230986743808,
%U 1209462292480,6332826779648,33159111507968,173623361929216,909103725543424,4760128905543680,24924358531088384,130505635564355584
%N Trinomial transform of the Fibonacci numbers (A000045).
%C Hankel transform of Sum_{k=0..n} (-1)^k*C(2k, k) (see A054108). - _Paul Barry_, Jan 13 2009
%C Hankel transform of A046748. - _Paul Barry_, Apr 14 2010
%C For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(2)'s along the three central diagonals. - _John M. Campbell_, Jul 12 2011
%C The limiting ratio is: Lim_{n -> oo} a(n)/a(n-1) = 1 + phi^3. - _Bob Selcoe_, Mar 18 2014
%C Invert transform of A052984. Invert transform is A083066. Binomial transform of A033887. Binomial transform is A163073. - _Michael Somos_, May 26 2014
%H Vincenzo Librandi, <a href="/A082761/b082761.txt">Table of n, a(n) for n = 0..150</a>
%H László Németh, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL21/Nemeth/nemeth6.html">The trinomial transform triangle</a>, J. Int. Seqs., Vol. 21 (2018), Article 18.7.3. Also <a href="https://arxiv.org/abs/1807.07109">arXiv:1807.07109</a> [math.NT], 2018.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-4).
%F a(n) = Sum_{k=0..2*n} A027907(n, k)*A000045(k+1).
%F From _Paul Barry_, Jul 16 2003: (Start)
%F Third binomial transform of (1, 1, 5, 5, 25, 25, ....).
%F a(n) = ((1+sqrt(5))(3+sqrt(5))^n-(1-sqrt(5))*(3-sqrt(5))^n)/(2*sqrt(5)). (End)
%F From _R. J. Mathar_, Nov 04 2008: (Start)
%F G.f.: (1-2*x)/(1-6*x+4*x^2).
%F a(n) = 6*a(n-1) - 4*a(n-2). (End)
%F a(n) = Sum_{k=0..n} A147703(n,k)*3^k. - _Philippe Deléham_, Nov 14 2008
%F For n>=2: a(n) = 5*a(n-1) + Sum_{i=1..n-2} a(i). - _Bob Selcoe_, Mar 18 2014
%F a(n) = a(-1-n) * 2^(2*n+1) for all n in Z. - _Michael Somos_, Mar 18 2014
%F a(n) = 2^n*Fibonacci(2*n+1), or 2^n*A001519(n+1). - _Bob Selcoe_, May 25 2014
%F From _Michael Somos_, May 26 2014: (Start)
%F a(n) - a(n-1) = A069429(n).
%F a(n+1) * a(n-1) - a(n)^2 = 4^n.
%F G.f.: 1 / (1 - 4*x / (1 - x / (1 - x))). (End)
%F E.g.f.: exp(3*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - _Stefano Spezia_, May 24 2024
%e a(5) = 2848 = 5*(544) + 4 + 20 + 104.
%e G.f. = 1 + 4*x + 20*x^2 + 104*x^3 + 544*x^4 + 2848*x^5 + 14912*x^6 + ...
%t a[ n_] := 2^n Fibonacci[ 2 n + 1]; (* _Michael Somos_, May 26 2014 *)
%t a[ n_] := If[ n < 0, SeriesCoefficient[ (2 - x) / (4 - 6 x + x^2), {x, 0, -1 - n}], SeriesCoefficient[ (1 - 2 x) / (1 - 6 x + 4 x^2), {x, 0, n}]]; (* _Michael Somos_, Oct 22 2017 *)
%t LinearRecurrence[{6,-4},{1,4},30] (* _Harvey P. Dale_, Jul 11 2014 *)
%o (PARI) a(n)=fibonacci(2*n+1)<<n \\ _Charles R Greathouse IV_, Jul 15 2011
%o (PARI) {a(n) = if( n<0, n = -1 - n; 2^(-1-2*n), 1) * polcoeff( (1 - 2*x) / (1 - 6*x + 4*x^2) + x * O(x^n), n)}; /* _Michael Somos_, Oct 22 2017 */
%o (Magma) [2^n * Fibonacci(2*n+1): n in [0..40]]; // _Vincenzo Librandi_, Jul 15 2011
%o (SageMath) [2^n*fibonacci(2*n+1) for n in range(41)] # _G. C. Greubel_, Jul 28 2024
%Y Cf. A000045, A001519, A027907, A033887, A046748, A052984, A054108, A069429, A083066, A147703, A163073.
%K easy,nonn
%O 0,2
%A _Emanuele Munarini_, May 21 2003