A Fib expression with multiplication

MattcAnderson · 2016-10-31, 23:56

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

science_man_88 · 2016-11-01, 00:50

Quote:

Originally Posted by MattcAnderson

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

MattcAnderson · 2016-11-01, 01:41

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

science_man_88 · 2016-11-01, 02:02

Quote:

Originally Posted by MattcAnderson

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

R. Gerbicz · 2016-11-01, 07:30

Quote:

Originally Posted by MattcAnderson

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).

MattcAnderson · 2016-11-01, 08:16

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

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Testing an expression for primality	1260	Software	17	2015-08-28 01:35
an expression parser for manual tests?	ixfd64	Software	3	2013-02-15 09:22
regular expression help	ixfd64	Programming	2	2009-03-01 06:19
Partial fraction of this expression?please!!	tinhnho	Miscellaneous Math	4	2005-01-17 19:45
Does Anyone Know how to Simplify the Following Expression?	jinydu	Puzzles	9	2004-04-02 01:03

2016-11-01, 01:41	#3
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 3·397 Posts	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

2016-11-01, 08:16	#6
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 1191₁₀ Posts	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(2n) = [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