Continued fractions as inverted recurrence relations

[Paul D. Hanna]

Joseph,
     Sorry I misspelled your name! 
Let me also finish the statement with Benoit Cloitre's formula for
A086377(n):
   
Given Joseph Biberstine's definition of b(n): 
b(n) = (n-1)^2/(b(n-1) - 2*n + 3) with b(1) = 4/Pi,
then 
round( b(n) ) = floor((1+sqrt(2))*n-1/sqrt(2)) = A086377(n).
 
I get that the first 190 terms agree - can someone check further?
And perhaps supply some theory as to why? 
      Paul


  
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