| Nearest integer to 1/6 * (2+sqrt(3))^(2n-1). - Ralf Stephan, Feb 24 2004
a(n) = A045899(n-1) + 1 = A051047(n+1) + 1 = A003697(2n-2). - N. J. A. Sloane, Jun 12 2004
a(n) = (A001835(n))^2. - Lekraj Beedassy, Jul 21 2006
Contribution from Paul Weisenhorn, May 12 2009: (Start)
With A=(2+sqrt(3))^2=7+4*sqrt(3) the equation x*x-3*m*m=1 has solutions
x(t)+sqrt(3)*m(t)=(2+sqrt(3))*A^t and the recurrences
x(t+2)=14*x(t+1)-x(t) with <x(t)> = 2,26,362,5042
m(t+2)=14*m(t+1)-m(t) with <m(t)> = 1,15,209,2911
a(t+2)=14*a(t+1)-a(t)-4 with <a(t)> = 1,9,121, as above. (End)
From Ant King, Nov 15 2011: (Start)
a(n) = 14*a(n-1) - a(n-2) - 4.
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (1/6)*( (2+sqrt(3))^(2n-1) + (2-sqrt(3))^(2n-1) + 2 ).
a(n) = ceiling( (1/6)*(2 + sqrt(3))^(2n-1) ).
a(n) = (1/6)*( (tan(5*pi/12))^(2n-1) + (tan(pi/12))^(2n-1) + 2 ).
a(n) = ceiling ( (1/6)*(tan(5*pi/12))^(2n-1) ).
G.f.: x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)). (End)
| 