Define x(n) + y(n)*sqrt(15) = (5+sqrt(15))*(4+sqrt(15))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+5*(s(n)^2s(n))).
From Richard Choulet, Sep 19 2007: (Start)
We must solve 3*p^2p=5*r^2+5*r+2, which gives X^2=15*Y^2+10 where X=6*p1 and Y=2*r+1.
Four other sequences are obtained at the same time:
X is given by 5,35,275,2165,... with the recurrence a(n+2)=8*a(n+1)a(n) and also a(n+1)=4*a(n)+(15*a(n)^2150)^(1/2) (numbers such that 15*X^2150 is a square).
Y is given by 1,9,71,559,... with the recurrence a(n+2)=8*a(n+1)a(n) and also a(n+1)=4*a(n)+(15*a(n)^2+10)^(1/2) (numbers such that 15*Y^2+10 is a square).
p is given by 1,6,46,361,... with the recurrence a(n+2)=8*a(n+1)a(n)1 and also a(n+1)=4*a(n)0.5+0.5*(60*a(n)^220*a(n)15)^(1/2) (numbers such that 15*(6*p1)^2150 is a square).
r is given by 0,4,35,279,... with the recurrence a(n+2)=8*a(n+1)a(n)+3 and also a(n+1)=4*a(n)+1.5+0.5*(60*a(n)^2+60*a(n)+25)^(1/2) (numbers such that 15*(2*r+1)^2+10 is a square).
a(n+2) = 62*a(n+1)a(n)10, a(n+1)=31*a(n)5+(960*a(n)^2320*a(n)45)^(1/2).
G.f.: ((z*(112*z+z^2)/((1z)*(162*z+z^2)).
(End)
a(n) = 1/6+(5/48)*sqrt(15)*[31+8*sqrt(15)]^(n1)(5/48)*sqrt(15)*[318*sqrt(15)]^(n1)+(5/12)*[31+8 *sqrt(15)]^(n1)+(5/12)*[318*sqrt(15)]^(n1), with n>=1.  Paolo P. Lava, Jul 15 2008
a(n) = 63*a(n1)63*a(n2)+a(n3).  Colin Barker, Jan 07 2015
