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
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).
Programs
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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
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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
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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
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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 *) -
PARI
my(x='x+O('x^30)); concat([0], Vec(x/(1-5*x+5*x^2))) \\ G. C. Greubel, Dec 27 2019 -
Sage
def A093131_list(prec): P.
= PowerSeriesRing(ZZ, prec) return P( x/(1-5*x+5*x^2) ).list() A093131_list(30) # G. C. Greubel, Dec 27 2019
Formula
G.f.: x/(1 - 5*x + 5*x^2).
a(n) = 5*a(n-1) - 5*a(n-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(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(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
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