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December 21st, 2012, 02:54 PM   #1
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Fibonacci

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December 21st, 2012, 04:04 PM   #2
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Re: Fibonacci

They're both linear recurrences ("C-finite"); the first is a(n) = 4a(n-1) + 3a(n-2) - 9a(n-3) + 2a(n-4) + a(n-5) and the second is a(n) = 5a(n-1) - 3a(n-2) - a(n-3).
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December 21st, 2012, 04:48 PM   #3
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Re: Fibonacci

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They're both linear recurrences ("C-finite"); the first is a(n) = 4a(n-1) + 3a(n-2) - 9a(n-3) + 2a(n-4) + a(n-5) and the second is a(n) = 5a(n-1) - 3a(n-2) - a(n-3).
Ok. For 2nd: by a(n) you mean a(3k); a=3k
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December 23rd, 2012, 02:22 PM   #4
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Re: Fibonacci

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Ok. For 2nd: by a(n) you mean a(3k); a=3k
I mean a(n) to be the sum of F(3k) from k = 1 to n. I don't mean a(3k).
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December 26th, 2012, 08:48 PM   #5
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Re: Fibonacci - I solved this...

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December 26th, 2012, 09:30 PM   #6
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Re: Fibonacci

And also
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December 27th, 2012, 05:20 AM   #7
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Re: Fibonacci

You might enjoy the book A = B, in which such identities are called "routine".
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December 27th, 2012, 07:27 AM   #8
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Re: Fibonacci

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You might enjoy the book A = B, in which such identities are called "routine".
I dont waste my time with useless stuff. I advice this to you, also.
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December 27th, 2012, 07:32 AM   #9
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Re: Fibonacci

What are you calling "useless", your identities or the book?
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December 27th, 2012, 07:48 PM   #10
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Re: Fibonacci

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What are you calling "useless", your identities or the book?
Book.
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