A093131 Binomial transform of Fibonacci(2n).

0, 1, 5, 20, 75, 275, 1000, 3625, 13125, 47500, 171875, 621875, 2250000, 8140625, 29453125, 106562500, 385546875, 1394921875, 5046875000, 18259765625, 66064453125, 239023437500, 864794921875, 3128857421875, 11320312500000, 40957275390625, 148184814453125

Offset

0,3

Comments

Second binomial transform of Fibonacci(n). - Paul Barry, Apr 22 2005

Links

Michael De Vlieger, Table of n, a(n) for n = 0..1791

G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.

S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.

M. Griffiths, Families of Sequences From a Class of Multinomial Sums, Journal of Integer Sequences, 15 (2012), #12.1.8.

László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.

J. Pan, Multiple Binomial Transforms and Families of Integer Sequences , J. Int. Seq. 13 (2010), 10.4.2, F^(2) and absolute values of F^(-2).

J. Pan, Some Properties of the Multiple Binomial Transform and the Hankel Transform of Shifted Sequences , J. Int. Seq. 14 (2011) # 11.3.4, remark 14.

Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).

Index entries for linear recurrences with constant coefficients, signature (5,-5).

Formula

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

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

a(n) = A093130(n)/2^n.

a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n, j)*C(j, k)*Fibonacci(j-k). - Paul Barry, Feb 15 2005

a(n) = Sum_{k=0..n} C(n, k)*2^k*Fibonacci(n-k) = Sum_{k=0..n} C(n, k)*2^(n-k) * Fibonacci(k). - Paul Barry, Apr 22 2005

a(n) = A030191(n-1), n > 0. - R. J. Mathar, Sep 05 2008

E.g.f.: 2*exp(5*x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Ilya Gutkovskiy, Aug 11 2017

From Kai Wang, Dec 22 2019: (Start)

a(n) = Sum_{i=0..n-1; j=0..n-1; i+2*j=n-1} 5^i*((i+j)!/(i!*j!)).

a(n*k)/a(k) = Sum_{i=0..n-1; j=0..n-1; i+2*j=n-1} (-1)^(j*(k-1))*b(k)^i*((i+j)!/(i!*j!)).

a((2*m+1)*k)/a(k) = Sum_{i=0..m-1} (-1)^(i*k)*A020876((2*m-2*i)*k) + 5^(m*k).

a(2*m*k)/a(k) = Sum_{i=0..m-1} (-1)^(i*k)*A020876((2*m-2*i-1)*k}.

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

a(m+r)*a(n+s) + a(m+s)*a(n+r) = (2*A020876(m+n+r+s) - 5^(n+s)*A020876(m-n)*A020876(r-s))/5.

A020876(m+r)*A020876(n+s) - A020876(m+s)*A020876(n+r) = 5^(n+s+1)*a(m-n)*a(r-s).

A020876(m+r)*A020876(n+s) - 5*a(m+s)*a(n+r) = 5^(n+s)*A020876(m-n)*A020876(r-s).

A020876(m+r)*A020876(n+s) + 5*a(m+s)*a(n+r) = 2*A020876(m+n+r+s) + 5^(n+s+1)*a(m-n)*a(r-s).

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

a(n)^2 - a(n+r)*a(n-r) = 5^(n-r)*a(r)^2.

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

a(m-n) = (a(m)*A020876(n) - A020876(m)*a(n))/(2*5^n).

a(m+n) = (a(m)*A020876(n) + A020876(m)*a(n))/2.

A020876(n)^2 - A020876(n+r)*A020876(n-r) = -5^(n-r+1)*a(r)^2.

A020876(m)*A020876(n+1) - A020876(m+1)*A020876(n) = -5^(n+1)*a(m-n).

A020876(m+n) - 5^(n)*A020876(m-n) = 5*a(m)*a(n).

A020876(m-n) = (A020876(m)*A020876(n) - 5*a(m)*a(n))/(2*5^n).

A020876(m+n) = (A020876(m)*A020876(n) + 5*a(m)*a(n))/2. (End)

a(2*n) = 5^n*Fibonacci(2*n), a(2*n+1) = 5^n*Lucas(2*n+1). - G. C. Greubel, Dec 27 2019

a(n) = Sum_{k=0..n} (-1)^(k+1)*binomial(2*n, n+k)*(k|5), where (k|5) is the Legendre symbol. - Greg Dresden, Oct 14 2022

Maple

seq(coeff(series(x/(1-5*x+5*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Dec 27 2019

Mathematica

CoefficientList[Series[x/(1-5x+5x^2), {x, 0, 30}], x] (* Michael De Vlieger, Dec 22 2019 *)
Table[If[EvenQ[n], 5^(n/2)*Fibonacci[n], 5^((n-1)/2)*LucasL[n]], {n, 0, 30}] (* G. C. Greubel, Dec 27 2019 *)
LinearRecurrence[{5, -5}, {0, 1}, 30] (* Harvey P. Dale, Mar 21 2023 *)

Prog

(PARI) my(x='x+O('x^30)); concat([0], Vec(x/(1-5*x+5*x^2))) \\ G. C. Greubel, Dec 27 2019
(Magma) I:=[0, 1]; [n le 2 select I[n] else 5*(Self(n-1) - Self(n-2)): n in [1..30]]; // G. C. Greubel, Dec 27 2019
(Sage)
def A093131_list(prec):
    P.<x> = PowerSeriesRing(ZZ, prec)
    return P( x/(1-5*x+5*x^2) ).list()
A093131_list(30) # G. C. Greubel, Dec 27 2019
(GAP) a:=[0, 1];; for n in [3..30] do a[n]:=5*(a[n-1]-a[n-2]); od; a; # G. C. Greubel, Dec 27 2019

Crossrefs

Cf. A000032, A000045, A020876, A030191.

Sequence in context: A092490 A094828 A030191 * A224422 A000344 A290922

Adjacent sequences: A093128 A093129 A093130 * A093132 A093133 A093134

Keyword

easy,nonn

Author

Paul Barry, Mar 23 2004

Status

approved