Bill, Pretty please, could you enter that yourself into sequence A001110 ? Go to https://oeis.org/A001110 Login (top right corner) - you are Bill Gosper (you can get a new passwiord at that point if you have forgotten it - it would be emailed to billgosper@gmail.com ) Then click edit, enter your updates etc, click Submit After 51 years of doing this myself, now we are a wiki, others can do it. More reliable and accurate , too! Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Tue, Jun 23, 2015 at 8:01 PM, Bill Gosper <billgosper@gmail.com> wrote:
Closed form (as square = triangular): ( (sqrt(2)+1)^(2*n)/(4*sqrt(2)) - (1-sqrt(2))^(2*n)/(4*sqrt(2)) )^2 = (1/2) * ( ( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2 + 1 )*( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2. - Bill Gosper <http://oeis.org/wiki/User:Bill_Gosper>, Jul 25 2008
And for the Mathematica section(?):
ChebyshevU[-1 + n, 3]^2 == Binomial[ChebyshevT[n/2, 3]^2, 2];
% /. n -> Range@9
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