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 2004-10-08, 04:48 #2 Zeta-Flux     May 2003 7×13×17 Posts Here are some possible lines of attack that I found. (You can plug them into LaTeX if you can't follow the notation). Define A_m = ( (1+\sqrt{2})^m + (1-\sqrt{2})^m )/2. Then this gets rid of the binomial stuff. And if we plug in m = k_n we get exactly the same numbers as you defined earlier. There are some interesting recurrence relations for the A_m. Look at: Code: A_1 = 1 > 0 A_2 = 1 > 2 > 2 > 0 A_3 = 3 > 2 > 4 > 4 > 4 >0 A_4 = 7 > 6 > 4 > 10 > 8 A_5 = 17 > 12 > 24 A_6 = 41 where the number immediately following > denotes the difference of the two previous numbers (to the left of >). Notice that every row is 2 times the row that occurs two places back. This pattern continues. So one can reconstruct this pattern using this fact and that the first part looks like: 1 > 0 Hope that gives you something new to ponder. (But I don't know if it will solve your problem.) Where did you get those k_n numbers from? Best, Zeta-Flux Last fiddled with by Zeta-Flux on 2004-10-08 at 04:51