%I #83 Sep 08 2022 08:44:45

%S 2,5,15,50,175,625,2250,8125,29375,106250,384375,1390625,5031250,

%T 18203125,65859375,238281250,862109375,3119140625,11285156250,

%U 40830078125,147724609375,534472656250,1933740234375,6996337890625,25312988281250,91583251953125

%N a(n) = ((5+sqrt(5))/2)^n + ((5-sqrt(5))/2)^n.

%C Number of no-leaf edge-subgraphs in Moebius ladder M_n.

%H Vincenzo Librandi, <a href="/A020876/b020876.txt">Table of n, a(n) for n = 0..1000</a>

%H Santiago Alzate, Oscar Correa, and Rigoberto Flórez, <a href="https://arxiv.org/abs/2009.02639">Fibonacci identities from Jordan Identities</a>, arXiv:2009.02639 [math.NT], 2020.

%H Noah Giansiracusa, <a href="https://doi.org/10.3390/math9040426">Fibonacci, Golden Ratio, and Vector Bundles</a>, Mathematics (2021) Vol. 9, Issue 4, 426.

%H J. P. McSorley, <a href="http://dx.doi.org/10.1016/S0012-365X(97)00086-1">Counting structures in the Moebius ladder</a>, Discrete Math., 184 (1998), 137-164.

%H Kai Wang, <a href="https://www.researchgate.net/publication/337943524_Fibonacci_Numbers_And_Trigonometric_Functions_Outline">Fibonacci Numbers And Trigonometric Functions Outline</a>, (2019).

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-5).

%F Also, a(n) = (sqrt(5)*phi)^n + (sqrt(5)/phi)^n, where phi = golden ratio. - _N. J. A. Sloane_, Aug 08 2014

%F Let S(n, m)=sum(k=0, n, binomial(n, k)*fibonacci(m*k)), then for n>0 a(n)= S(2*n, 2)/S(n, 2). - _Benoit Cloitre_, Oct 22 2003

%F From _R. J. Mathar_, Feb 06 2010: (Start)

%F a(n)= 5*a(n-1) - 5*a(n-2).

%F G.f.: (2-5*x)/(1-5*x+5*x^2). (End)

%F From _Johannes W. Meijer_, Jul 01 2010: (Start)

%F Lim_{k->infinity} a(n+k)/a(k) = (A020876(n) + A093131(n)*sqrt(5))/2.

%F Lim_{k->infinity} A020876(n)/A093131(n) = sqrt(5). (End)

%F Binomial transform of A005248. - _Carl Najafi_, Sep 10 2011

%F a(n) = 2*A030191(n) - 5*A030191(n-1). - _R. J. Mathar_, Mar 02 2012

%F From _Kai Wang_, Dec 22, 2019: (Start)

%F a((2*m+1)*k)/a(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*a(2*(m-i)*k) + 5^(m*k).

%F A093131(m+r)*A093131(n+s) + A093131(m+s)*A093131(n+r) = (2*a(m+n+r+s) - 5^(n+s)*a(m-n)*a(r-s))/5.

%F a(m+r)*a(n+s) - a(m+s)*a(n+r) = 5^(n+s+1)*A093131(m-n)*A093131(r-s).

%F a(m+r)*a(n+s) + a(m+s)*a(n+r) = 2*a(m+n+r+s) + 5^(n+s)*a(m-n)*a(r-s).

%F a(m+r)*a(n+s) - 5*A093131(m+s)*A093131(n+r) = 5^(n+s)*a(m-n)*a(r-s).

%F a(m+r)*a(n+s) + 5*A093131(m+s)*A093131(n+r) = 2*a(m+n+r+s)+ 5^(n+s+1)*A093131(m-n)*A093131(r-s).

%F A093131(m-n) = (A093131(m)*a(n) - a(m)*A093131(n))/(2*5^n).

%F A093131(m+n) = (A093131(m)*a(n) + a(m)*A093131(n))/2.

%F a(n)^2 - a(n+1)*a(n-1) = -5^n.

%F a(n)^2 - a(n+r)*a(n-r) = -5^(n-r+1)*A093131(r)^2.

%F a(m)*a(n+1) - a(m+1)*a(n) = -5^(n+1)*A093131(m-n).

%F a(m+n) - 5^(n)*a(m-n) = 5*A093131(m)*A093131(n).

%F a(m+n) + 5^(n)*a(m-n) = a(m)*a(n).

%F a(m-n) = (a(m)*a(n) - 5*A093131(m)*A093131(n))/(2*5^n).

%F a(m+n) = (a(m)*a(n) + 5*A093131(m)*A093131(n))/2. (End)

%F E.g.f.: 2*exp(5*x/2)*cosh(sqrt(5)*x/2). - _Stefano Spezia_, Dec 27 2019

%e G.f. = 2 + 5*x + 15*x^2 + 50*x^3 + 175*x^4 + 625*x^5 + 2250*x^6 + ...

%p G:=(x,n)-> cos(x)^n+cos(3*x)^n:

%p seq(simplify(4^n*G(Pi/10,2*n)), n=0..22); # _Gary Detlefs_, Dec 05 2010

%t Table[Sum[LucasL[2*i] Binomial[n, i], {i, 0, n}], {n, 0, 50}] (* _T. D. Noe_, Sep 10 2011 *)

%t CoefficientList[Series[(2 - 5 x)/(1 - 5 x + 5 x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Aug 08 2014 *)

%t LinearRecurrence[{5,-5},{2,5},30] (* _Harvey P. Dale_, Mar 13 2016 *)

%o (Sage) [lucas_number2(n,5,5) for n in range(0,24)] # _Zerinvary Lajos_, Jul 08 2008

%o (Magma) [Floor(((5+Sqrt(5))/2)^n+((5-Sqrt(5))/2)^n): n in [0..30]]; // _Vincenzo Librandi_, Aug 08 2014

%Y Cf. A020876, A093131, A005248.

%Y Appears in A109106. - _Johannes W. Meijer_, Jul 01 2010

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_

%E Definition simplified by _N. J. A. Sloane_, Aug 08 2014