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URL: https://oeis.org/A093131

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A093131
Binomial transform of Fibonacci(2n).
8
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
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).
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(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
(SageMath)
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
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Mar 23 2004
STATUS
approved