The identity (648*n^2-72*n+1)^2-(9*n^2-n)*(216*n-12)^2=1 can be written as A154514(n)^2-a(n)*A154518(n)^2=1 (see also the second comment in A154514). - Vincenzo Librandi, Jan 30 2012
The continued fraction expansion of sqrt(a(n)) is [3n-1; {1, 4, 1, 6n-2}]. For n=1, this collapses to [2; {1, 4}]. - Magus K. Chu, Sep 06 2022