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 March 17th, 2009, 04:02 PM #1 Newbie   Joined: Mar 2009 Posts: 5 Thanks: 0 What is the next number in the sequence? 2, 2, 8, 72, 392, 2312, ?
 March 17th, 2009, 05:35 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: What is the next number in the sequence? Dividing the sequence term-by-term by 2 gives 1,1,4,36,196,1156 which is $1^2,1^2,2^2,6^2,14^2,34^2.$ The sequence 1,2,6,14,34 can be generated by the expression $x_{n+1}=2x_n+x_{n-1},$ which leads me to believe that either the original sequence should start with only a single 2, or the first 2 should be a 0. Assuming this is the case, the next term in the 1,2,6,14,34 sequence will be (2 x 34) + 14;I'm sure you can work out the rest yourself.
 March 17th, 2009, 09:34 PM #3 Newbie   Joined: Mar 2009 Posts: 5 Thanks: 0 Re: What is the next number in the sequence? You are absolutely correct! Amazing! I think you are a genius for figuring that out. The sequence is 2, 8, 72, 392, 2312, 13448, ... I would never have figured out that they would all have even square roots if you divide them all by 2. The equation would be: $sqrt{x_{n+1}/2}=2sqrt{x_n/2}+sqrt{x_{n-1}/2}$ Now for the cool part. Check out the attached image. The first 2 rectangles each have an area of 2. The next largest has an area of 8, and so on, and so on, and so on.... Sorry about the 2 issue. I wasn't sure how to write the beginning of the sequence since I didn't know the equation.
 March 18th, 2009, 11:40 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: What is the next number in the sequence? I should actually mention that the method of producing that sequence only works for 2,2,6,14,34,... , which is presumably because the diagram could actually include infinitely many small rectangles, so the first terms of the sequence need to be artificially tweaked. Nevertheless, it is possible to find an explicit formula for the sequence 2,6,14,34,...; it is actually $x_n=(1+\sqrt2)^n+(1-\sqrt2)^n,\qquad n\geq1.$ The sequence 8,72,392,2312,... can then be expressed by $y_n=2{x_n}^2=2\left((1+\sqrt2)^{2n}+(1-\sqrt2)^{2n}+2(-1)^n\right).$
March 18th, 2009, 01:59 PM   #5
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Re: What is the next number in the sequence?

Or the sequence 8,8,72,392,2312,... can be expressed by
Quote:
 $y_n=2{x_n}^2=2\left((1+\sqrt2)^{2n}+(1-\sqrt2)^{2n}+2(-1)^n\right).$
hmm...

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