For positive n, a(n) equals [1,1;1,3]^n.[1,2].[1.2], which means calculate the nth power of the specific 2x2 matrix, multiply from the right by the column vector [1,2], and finally take the dot product of this column vector with the vector [1,2]. [From John M. Campbell, Jul 09 2011]
a(n) is the 3rd subdiagonal of array A228405.  Richard R. Forberg, Sep 02 2013
From Max Alekseyev, Sep 03 2013: (Start)
It is not hard to show that A010914 indeed satisfies the following simple linear recurrence relation: a(n) = 4*a(n1)  2*a(n2).
Let a(n) = A010914(n) be the sequence defined by the recurrence
a(n)=floor( a(n1)^2/a(n2)+1/2 ) and initial terms a(0)=5 and a(1)=17.
Define a sequence b(n) by the recurrence b(n) = 4*b(n1)  2*b(n2) and the same initial terms b(0)=a(0)=5 and b(1)=a(1)=17.
We want to prove that a(n)=b(n) for all n.
The main trick is that instead of proving that the sequence a() satisfies the recurrence for the sequence b(), we will do it the other way around: prove that b() satisfied the recurrence for a(). This will imply that these two sequences coincide.
To prove that b() satisfies the recurrence for a(), it is enough to show that abs( b(n1)^2  b(n)*b(n2) ) < b(n2)/2.
To do this, we will use the explicit formula for b(n):
b(n) = (10+7*sqrt(2))/4 * (2+sqrt(2))^n + (107*sqrt(2))/4 * (2sqrt(2))^n
Explicitly computing abs( b(n1)^2  b(n)*b(n2) ), one finds that it equals 2^n times a constant. At the same time, b(n) grows as (2+sqrt(2))^n (up to a constant) which allows one to easily prove (e.g., by induction on n) that abs( b(n1)^2  b(n)*b(n2) ) < b(n2)/2. (End)
a(n) = A079496(n+3) * A016116(n).  Michael Somos, Sep 03 2013
a(n)  a(n1)^2 / a(n2) = 2^floor(n/2) / A143609(n+1) < 1/2 if n>1.  Michael Somos, Sep 03 2013
