Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fibonacci Numbers - out of curiousity

  1. Apr 5, 2010 #1
    This are some fibonacci numbers:


    Such that

    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 -

    Attached Files:

    • eqn.jpg
      File size:
      8.2 KB
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 5, 2010 #2
    Shouldn't those more appropriately be called the Pingala numbers?
  4. Apr 5, 2010 #3
    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...
  5. Apr 5, 2010 #4
    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).
  6. Apr 5, 2010 #5
    Oh. I didn't know that . I was always been told that the sequence of numbers are called fibonacci numbers. =x Thanks btw
  7. Apr 5, 2010 #6


    Staff: Mentor

    What's the equation? The attachment is still waiting to be approved.
    Last edited by a moderator: May 4, 2017
  8. Apr 5, 2010 #7
  9. Apr 6, 2010 #8
    [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]
  10. Jul 23, 2011 #9

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook