"A Diofloortin equation": n such that 2*n^2=floor(n*sqrt(2)*ceiling(n*sqrt(2))).
a(n)*a(n+3) = -2 + a(n+1)*a(n+2).
G.f.: x*(1+x)^2/(1-6*x^2+x^4);
a(n) = ((sqrt(2)+1)^n-(sqrt(2)-1)^n)*((sqrt(2)/8-1/4)*(-1)^n+sqrt(2)/8+1/4);
a(n) = Sum_{k=0..floor(n/2)} 2^k*(C(n,2*k)-C(n-1,2*k+1)*(1+(-1)^n)/2). (End)
a(2*n + 2) = a(2*n + 1) + sqrt( (1 + a(2*n + 1)^2)/2 ).
a(2*n + 1) = 2*a(2*n) + sqrt( (1 + 2*a(2*n)^2) ).
More generally,
a(2*n+2*m+1) = sqrt(2)*a(2*n) o a(2*m+1), where o is the binary operation defined above, that is,
a(2*n+2*m+1) = sqrt(2)*a(2*n)*sqrt(1 + a(2*m+1)^2) + a(2*m+1)*sqrt(1 + 2*a(2*n)^2).
sqrt(2)*a(2*(n + m)) = (sqrt(2)*a(2*n)) o (sqrt(2)*a(2*m)), that is,
a(2*n+2*m) = a(2*n)*sqrt(1 + 2*a(2*m)^2) + a(2*m)*sqrt(1 + 2*a(2*n)^2).
(1 + a(2*n+1)^2)/2 =
A008844(n). (End)