G.f.: (1-2*x)/(1-3*x+x^2).
G.f.: 1 / (1 - x / (1 - x / (1 - x))). -
Michael Somos, May 03 2012
a(n) = a(1-n) for all n in Z.
a(n+2) = (a(n+1)^2+1)/a(n) with a(1)=1, a(2)=2. -
Benoit Cloitre, Aug 29 2002
a(n) = (phi^(2*n-1) + phi^(1-2*n))/sqrt(5) where phi=(1+sqrt(5))/2. -
Michael Somos, Oct 28 2002
a(n) = Sum_{k=0..n} binomial(n+k, 2*k). -
Len Smiley, Dec 09 2001
a(n) ~ (1/5)*sqrt(5)*phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 1)=a(n) (this comment is essentially the same as that of L. Smiley). -
Benoit Cloitre, Nov 10 2002
a(n) = (1/2)*(3*a(n-1) + sqrt(5*a(n-1)^2-4)). -
Benoit Cloitre, Apr 12 2003
Main diagonal of array defined by T(i, 1) = T(1, j) = 1, T(i, j) = max(T(i-1, j) + T(i-1, j-1); T(i-1, j-1) + T(i, j-1)). -
Benoit Cloitre, Aug 05 2003
Solutions x > 0 to equation floor(x*r*floor(x/r)) = floor(x/r*floor(x*r)) when r=phi. -
Benoit Cloitre, Feb 15 2004
a(n) = Sum_{i=0..n} binomial(n+i, n-i). -
Jon Perry, Mar 08 2004
a(n) = S(n-1, 3) - S(n-2, 3) = T(2*n-1, sqrt(5)/2)/(sqrt(5)/2) with S(n, x) = U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first kind. See triangle
A049310, resp.
A053120. -
Wolfdieter Lang, Aug 31 2004
a(n) = ((-1)^(n-1))*S(2*(n-1), i), with the imaginary unit i and S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind,
A049310. -
Wolfdieter Lang, Aug 31 2004
a(n) = Sum_{0<=i_1<=i_2<=n} binomial(i_2, i_1)*binomial(n, i_1+i_2). -
Benoit Cloitre, Oct 14 2004
a(n) = a(n-1) + Sum_{i=0..n-1} a(i)*a(n) = F(2*n+1)*Sum_{i=0..n-1} a(i) = F(2*n). - Andras Erszegi (erszegi.andras(AT)chello.hu), Jun 28 2005
The i-th term of the sequence is the entry (1, 1) of the i-th power of the 2 X 2 matrix M = ((1, 1), (1, 2)). -
Simone Severini, Oct 15 2005
a(n-1) = (1/n)*Sum_{k=0..n} B(2*k)*F(2*n-2*k)*binomial(2*n, 2*k) where B(2*k) is the (2*k)-th Bernoulli number. -
Benoit Cloitre, Nov 02 2005
a(n) = (2/sqrt(5))*cosh((2n-1)*psi), where psi=log(phi) and phi=(1+sqrt(5))/2. -
Hieronymus Fischer, Apr 24 2007
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3); a(n) = ((sqrt(5) + 5)/10)*(3/2 + sqrt(5)/2)^(n-2) + ((-sqrt(5) + 5)/10)*(3/2 - sqrt(5)/2)^(n-2). -
Antonio Alberto Olivares, Mar 21 2008
With X,Y defined as X = ( F(n) F(n+1) ), Y = ( F(n+2) F(n+3) ), where F(n) is the n-th Fibonacci number (
A000045), it follows a(n+2) = X.Y', where Y' is the transpose of Y (n >= 0). -
K.V.Iyer, Apr 24 2009
a(n) = Fibonacci(2*n+2) mod Fibonacci(2*n), n > 1.
a(n) = (Fibonacci(n-1)^2 + Fibonacci(n)^2 + Fibonacci(2*n-1))/2. (End)
a(n) = 2^n*f(n;1/2), where f(n;d), n=0,1,...,d, denote the so-called delta-Fibonacci numbers (see Witula et al. papers and comments in
A000045). -
Roman Witula, Jul 12 2012
a(n) = (Fibonacci(n+2)^2 + Fibonacci(n-3)^2)/5. -
Gary Detlefs, Dec 14 2012
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 - x/(x*k + 1 )/Q(k+1); (recursively defined continued fraction). -
Sergei N. Gladkovskii, Feb 23 2013
G.f.: (1-2*x)*G(0)/(2-3*x), where G(k) = 1 + 1/( 1 - x*(5*k-9)/(x*(5*k-4) - 6/G(k+1) )); (continued fraction). -
Sergei N. Gladkovskii, Jul 19 2013
G.f.: 1 + x*(1-x^2)*Q(0)/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 + 2*x - x^2)/( x*(4*k+4 + 2*x - x^2 ) + 1/Q(k+1) )); (continued fraction). -
Sergei N. Gladkovskii, Sep 11 2013
G.f.: Q(0,u), where u=x/(1-x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k+1 + u)/Q(k+1); (continued fraction). -
Sergei N. Gladkovskii, Oct 07 2013
Let F(n) be the n-th Fibonacci number,
A000045(n), and L(n) be the n-th Lucas number,
A000032(n). Then for n > 0, a(n) = F(n)*L(n-1) + (-1)^n. -
Charlie Marion, Jan 01 2014
1 = a(n)*a(n+2) - a(n+1)*a(n+1) for all n in Z. -
Michael Somos, Jul 08 2014
a(n) = 3*F(n-1)^2 + F(n-3)*F(n) - 2*(-1)^n. -
J. M. Bergot, Feb 17 2016
E.g.f.: (2*exp(sqrt(5)*x) + 3 + sqrt(5))*exp(-x*(sqrt(5)-3)/2)/(5 + sqrt(5)). -
Ilya Gutkovskiy, Jul 04 2016
a(n) = ((M_2)^n)[1,1] = S(n, 3) - 2*S(n-1, 3), with the 2 X 2 tridiagonal matrix M_2 = Matrix([1,1], [1,2]) from
A322602. For a proof see the Mar 30 2020 comment above. -
Wolfdieter Lang, Mar 30 2020
a(n+1) = Product_{k=1..n} (1 + 4*cos(2*Pi*k/(2*n + 1))^2). Special case of
A099390. -
Greg Dresden, Oct 16 2021
a(n+1) = 4^(n+1)*Sum_{k >= n} binomial(2*k,2*n)*(1/5)^(k+1). Cf.
A102591. -
Peter Bala, Nov 29 2021
a(n) = cosh((2*n-1)*arcsinh(1/2))/sqrt(5/4). -
Peter Luschny, May 21 2022
a(n) = F(n-1)*L(n) - (-1)^n where L(n)=
A000032(n) and F(n)=
A000045(n).
a(n) = (L(n-1)^2 + L(n-1)*L(n+1))/5 + (-1)^n.
a(n) = 2*(area of a triangle with vertices at (L(n-2), L(n-1)), (F(n), F(n-1)), (L(n), L(n+1))) + 5*(-1)^n for n > 2. (End)
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