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From Eric Simon Jacob, Jul 20 2023: (Start)

G.f.: (1 + x - 3*x^2 - 2*x^3) / (x^4 - 4*x^2 + 1).

a(n) = 0*a(n-1) + 4*a(n-2) + 0*a(n-3) - 1*a(n-4) for n > 3, a(0) = 1, a(1) = 1, a(2) = 1, a(3) = 2.

A = (989*sqrt(3)+1713-(2340*sqrt(3)+4053)*sqrt(2-sqrt(3)))/(9360*sqrt(3)+16212)

B = ((45*sqrt(3)+78)*sqrt(2-sqrt(3))+19*sqrt(3)+33)/(180*sqrt(3)+312)

C = ((2*sqrt(3)+3)*sqrt(2+sqrt(3))+3*sqrt(3)+5)/(8*sqrt(3)+12)

D = (3*sqrt(3)+5-(2*sqrt(3)+3)*sqrt(2+sqrt(3)))/(8*sqrt(3)+12)

A ~ -0.0237470898486668

B ~ 0.235071955253854

C ~ 0.877300480441941

D ~ -0.0886253458471277

a(n) = ((-1)^n*A + B)*sqrt(2+sqrt(3))^n + (C + (-1)^n*D)*sqrt(2-sqrt(3))^n.

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For n > 2, a(n) is the smallest integer > a(n-1) such that sqrt(3)*a(n) is closer to and greater than an integer than sqrt(3)*a(n-1). I.e., a(n) is the smallest integer > a(n-1) such that frac(sqrt(3)*a(n)) < frac(sqrt(3)*a(n-1)). - Benoit Cloitre, Jan 20 2003

This sequence is a particular case of the following situation: a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n) where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z. The g.f . is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4); so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n). The general form of a(n) is given by: a(2*m) = Sum_{p=sum0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p), + (b-Q)*Sum_{p=0..floor((m-1)/2))+(b-Q)*sum(} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p), and a(2*m+1) = a*Sum_{p=0..floor((m-1)/2)) and a(2*m+1)=a*sum(} (-1)^p*binomial(m-p,p)*Q^(m-2*p), + (a*b+q-a*Q)*Sum_{p=0..floor((m-1)/2))+(a*b+q-a*Q)*sum(} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)). - Richard Choulet, Feb 24 2010

T. D. Noe, <a href="/A005246/b005246.txt">Table of n, a(n) for n = 0..500</a>

Limit_{n->infinityoo} a(2n+1)/a(2n) = (3+sqrt(3))/3 = 1.5773502...; lim_{n->infinityoo} a(2n)/a(2n-1) = (3+sqrt(3))/2 = 2.3660254.... - Benoit Cloitre, Aug 07 2002

a(n) = a(2 - n) for all n in Z. - Michael Somos, Nov 15 2006

For n > 2: a(n) = a(n-1) + Sum_{k=1..floor((n-1)/2)} a(2*k). - Reinhard Zumkeller, Dec 16 2007

In the closed form formula (sqrt(2+sqrt(3))^n)=((sqrt(6)+sqrt(2))/2)^n;(-sqrt(2+sqrt(3))^n)=((-sqrt(6)-sqrt(2))/2)^n;(sqrt(2-sqrt(3))^n)=((sqrt(6)-sqrt(2))/2)^n ;(-sqrt(2-sqrt(3))^n)=((sqrt(2)-sqrt(6))/2)^n. - Tim Monahan, Jul 07 2011

From Tim Monahan, Jul 07 2011: (Start)

In the closed-form formula,

sqrt(2+sqrt(3))^n = ((sqrt(6)+sqrt(2))/2)^n;

-sqrt(2+sqrt(3))^n = ((-sqrt(6)-sqrt(2))/2)^n;

sqrt(2-sqrt(3))^n = ((sqrt(6)-sqrt(2))/2)^n;

-sqrt(2-sqrt(3))^n = ((sqrt(2)-sqrt(6))/2)^n.

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Integer solutions to the equation floor(sqrt(3)*x^2) = x*floor(sqrt(3)*x). - Benoit Cloitre, Mar 18 2004

For n > 2, a(n) is the smallest integer > a(n-1) such that sqrt(3)*a(n) is closer to and greater than an integer than sqrt(3)*a(n-1). iI.e. , a(n) is the smallest integer > a(n-1) such that frac(sqrt(3)*a(n))<frac(sqrt(3)*a(n-1)). - Benoit Cloitre, Jan 20 2003

a(n) for n > 1 are the integer square roots of (floor(m^2/3)+1), where the values of m are given by A143643. Also see A082630. - Richard R. Forberg, Nov 14 2013

a(n) = 0*a(n-1) + 4*a(n-2) + 0*a(n-3) - 1*a(n-4) for n > 3, a(0) = 1, a(1) = 1, a(2) = 1, a(3) = 2.

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a(n) = 0.*a(n-1) + 4.*a(n-2) + 0.*a(n-3) -1.*a(n-4) for n>3, a(0) = 1, a(1) = 1, a(2) = 1, a(3) = 2.