20041006, 21:07  #1 
Feb 2004
France
2^{2}·229 Posts 
I need a proof for this binomial property.
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 
20041008, 04:48  #2 
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 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, ZetaFlux Last fiddled with by ZetaFlux on 20041008 at 04:51 
20041008, 15:53  #3 
Feb 2004
France
1110010100_{2} Posts 
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_{n1}+X_{n2} 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_{n1} = 4 U_{n1} or 4 U_n = V_n + V_{n1} . 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 
20041008, 19:13  #4 
May 2003
11000001011_{2} Posts 
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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