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#1 |
Feb 2004
France
3·311 Posts |
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Hi,
Because the following forum provides a LaTeX interface and because the formula is quite complex, I've posted a thread in the NumberTheory forum at: http://www.physicsforums.com/showthread.php?t=46414 asking for help for proving a property about binomial coefficients I need. Your help is welcome ! (And I guess finding a proof will not be easy). Regards, Tony |
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#2 |
May 2003
7·13·17 Posts |
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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 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 |
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#3 |
Feb 2004
France
3·311 Posts |
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Thanks.
But I'm not sure it helps. In fact, your operation seems to periodically (period = 2) build the same 2 series, but shifted to the bottom by 1 line. Add some lines to your table, and you'll see 2 series: U_n : 0 1 2 5 12 29 ... V_n : 2 2 6 14 34 82 ... X_n = 2X_{n-1}+X_{n-2} which are the Pell sequences. And on the right, you'll see: 1 0 2 0 4 0 8 0 ... 0 2^i ... So we have: V_n - V_{n-1} = 4 U_{n-1} or 4 U_n = V_n + V_{n-1} . I don't know yet if it helps. Using the binomial stuff is one solution for studying the problem. Using the relationship between Pell numbers is another one. Don't know which will provide the proof ... About the k_n, I'm writing a paper that will explain everything, soon. In fact: F_n is prime <==> F_n | A_k_n , I think. Tony |
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#4 |
May 2003
60B16 Posts |
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You are exactly right. That's what I was trying to say. (Sorry about the mistaken 12 instead of 14 in my table.)
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