Fibonacci Numbers - out of curiousity

  • Thread starter icystrike
  • Start date
  • #1
446
1

Main Question or Discussion Point

This are some fibonacci numbers:

0,1,1,2,3,5,...

Such that
[URL]http://upload.wikimedia.org/math/0/c/e/0cebc512d9a3ac497eda6f10203f792e.png[/URL]

with seed values
[PLAIN]http://upload.wikimedia.org/math/9/e/4/9e47cfc063b09aeee2c39ad594759dd8.png [Broken] [Broken] and [PLAIN]http://upload.wikimedia.org/math/9/e/4/9e47cfc063b09aeee2c39ad594759dd8.png [Broken] [Broken]

Find the greatest integer n that satisfy the below equation.
p.s : That should be a + instead of -
attachment.php?attachmentid=24896&stc=1&d=1270482112.jpg
 

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Answers and Replies

  • #2
69
0
Shouldn't those more appropriately be called the Pingala numbers?
 
  • #3
446
1
hmm.. would you tell me moer about it?
I was just trying to apply what i've learned in introductory number theory , linear algebra and calculus into fibonacci numbers...
 
  • #4
69
0
I seem to remember somebody telling me that Fibonacci published his "Liber Abaci" in 1202, but that his famous sequence was actually plagiarised form the Indian mathematician Pingala (as was common at that time).
 
  • #5
446
1
Oh. I didn't know that . I was always been told that the sequence of numbers are called fibonacci numbers. =x Thanks btw
 
  • #6
33,496
5,188
This are some fibonacci numbers:

0,1,1,2,3,5,...

Such that
[URL]http://upload.wikimedia.org/math/0/c/e/0cebc512d9a3ac497eda6f10203f792e.png[/URL]

with seed values
[PLAIN]http://upload.wikimedia.org/math/9/e/4/9e47cfc063b09aeee2c39ad594759dd8.png [Broken] [Broken] and [PLAIN]http://upload.wikimedia.org/math/9/e/4/9e47cfc063b09aeee2c39ad594759dd8.png [Broken] [Broken]

Find the greatest integer n that satisfy the below equation.
p.s : That should be a + instead of -
attachment.php?attachmentid=24896&stc=1&d=1270482112.jpg
What's the equation? The attachment is still waiting to be approved.
 
Last edited by a moderator:
  • #8
446
1
[tex]\lfloor\frac{(\frac{(F_{m}}{F_{m-1})^{k}}{2\times\frac{(F_{m}}{F_{m-1})-1}+\frac{1}{2}\rfloor = F_{k} [/tex]
 
  • #9
26
0
Fn=([itex]\phi[/itex]n-(1-[itex]\phi[/itex])n)/[itex]\sqrt{}[/itex]5

If Fn=[itex]\left\lfloor[/itex](4/9)(13/8)n-1/2[itex]\right\rfloor[/itex] then 9(Fn-1/2)/4>(13/8)n. Since 13/8>[itex]\phi[/itex], the right hand side increases faster, and the right hand side for n=10 is already greater than the left hand side. 8 and 9 are the only integer solutions.
 

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