20161031, 23:56  #1 
"Matthew Anderson"
Dec 2010
Oregon, USA
601 Posts 
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 
20161101, 00:50  #2 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
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
Last fiddled with by science_man_88 on 20161101 at 01:17 
20161101, 01:41  #3 
"Matthew Anderson"
Dec 2010
Oregon, USA
1131_{8} 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 
20161101, 02:02  #4  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
Last fiddled with by science_man_88 on 20161101 at 02:13 

20161101, 07:30  #5 
"Robert Gerbicz"
Oct 2005
Hungary
2×3×5×47 Posts 
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(m1)*F(n1)=F(m+n1). 
20161101, 08:16  #6 
"Matthew Anderson"
Dec 2010
Oregon, USA
601 Posts 
Hi Math and computer people,,
Thank you Mr. R. Gerbicz for your comment. Assuming that F(n) = F(n1) + F(n2) implies that From the Wikipedia article about Fibonacci numbers https://en.wikipedia.org/wiki/Fibonacci_number I quote  F(m)*F(n) + F(m1)*F(n1) = F(m+n1), F(m)*F(n+1) + F(m1)*F(n) = F(m+n) In particular, with m=n we infer that F(2n1) = F(n)^2 + F(n1)^2 and F(2*n) = [F(n1) + 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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