a(n) = 5*a(n-1) - a(n-2) + 2, a(0)=1, a(1)=7.
a(n) = (1/3)*(-2 + ((5+sqrt(21))/2)^n + ((5-sqrt(21))/2)^n). -
Ralf Stephan, Apr 14 2004
G.f.: (1+x)/((1-x)*(1 - 5*x + x^2)) = (1+x)/(1 - 6*x + 6*x^2 - x^3). From the R. Stephan link.
a(n) = 6*a(n-1) - 6*a(n-2) + a(n-3), n>=2, a(-1):=0, a(0)=1, a(1)=7.
a(n) = (2*T(n, 5/2)-2)/3, with twice the Chebyshev polynomials of the first kind, 2*T(n, x=5/2)=
A003501(n).
a(n) = b(n) + b(n-1), n>=1, with b(n)=
A089817(n) the partial sums of S(n, 5)= U(n, 5/2)=
A004254(n+1), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind.
The following formulas assume an offset of 1.
Let {u(n)} be the Lucas sequence in the quadratic integer ring Z[sqrt(7)] defined by the recurrence u(0) = 0, u(1) = 1 and u(n) = sqrt(7)*u(n-1) - u(n-2) for n >= 2. Then a(n) = u(n)^2.
Equivalently, a(n) = U(n-1,sqrt(7)/2)^2, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = 1/3*( ((sqrt(7) + sqrt(3))/2)^n - ((sqrt(7) - sqrt(3))/2)^n )^2.
a(n) = bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -5/2; 1, 7/2] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in
A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
0 = 1 + a(n)*(-2 + a(n) - 5*a(n+1)) + a(n+1)*(-2 + a(n+1)) for all n in Z. -
Michael Somos, Jan 22 2017
