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Old 2016-10-31, 23:56   #1
MattcAnderson
 
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"Matthew Anderson"
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Default A Fib expression with multiplication

Hi Math and Computer people,

Some of you may be familiar with the standard Fibonacci expression,, definition.

This is a request for comments.

Please let me know if there are errors.

Regards,
Matt
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Old 2016-11-01, 00:50   #2
science_man_88
 
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Quote:
Originally Posted by MattcAnderson View Post
Hi Math and Computer people,

Some of you may be familiar with the standard Fibonacci expression,, definition.

This is a request for comments.

Please let me know if there are errors.

Regards,
Matt
the only logical error is you said it didn't have a start but then by naming the coefficient starting within the sequence itself you gave it a start at the second 1 in the sequence of 0,1,1,2,3,5,8 ... it's an example of a lucas sequence ( not to be confused with the lucas numbers). edit: doh and one part you say 1 1 2 3 8 instead of 1 2 3 5 8

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Old 2016-11-01, 01:41   #3
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Hi again,

Thank you ScienceMan88. You make a good point. There is definitely a difference between the Fibonacci sequence and the Lucas sequence.

Regards,
Matt
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Old 2016-11-01, 02:02   #4
science_man_88
 
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Quote:
Originally Posted by MattcAnderson View Post
Hi again,

Thank you ScienceMan88. You make a good point. There is definitely a difference between the Fibonacci sequence and the Lucas sequence.

Regards,
Matt
terminology counts a lot in sciences as this forum proves again and again you'll find lucas numbers is the correct term the fibonacci numbers technically are a lucas sequence

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Old 2016-11-01, 07:30   #5
R. Gerbicz
 
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Quote:
Originally Posted by MattcAnderson View Post
This is a request for comments.
Probably this is an 800 years old formula. See https://en.wikipedia.org/wiki/Fibonacci_number at "Matrix form" section, with different letters:
F(m)*F(n)+F(m-1)*F(n-1)=F(m+n-1).
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Old 2016-11-01, 08:16   #6
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Hi Math and computer people,,
Thank you Mr. R. Gerbicz for your comment.
Assuming that
F(n) = F(n-1) + F(n-2)
implies that
From the Wikipedia article about Fibonacci numbers
https://en.wikipedia.org/wiki/Fibonacci_number
I quote -
F(m)*F(n) + F(m-1)*F(n-1) = F(m+n-1),
F(m)*F(n+1) + F(m-1)*F(n) = F(m+n)
In particular, with m=n we infer that
F(2n-1) = F(n)^2 + F(n-1)^2
and
F(2*n) = [F(n-1) + F(n+1)]*F(n).

This expression is available in the literature.

At least it is less than 1000 years old. *sigh*
Regards,
Matt
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