User talk:Joseph P. Shoulak

When he was a young boy, Leonard Bernstein figured out all the principles of harmony (major and minor chords, dominant sevenths, etc.) by himself. He was very smart, but he hadn't discovered anything new. A lot of what he figured out on his own had been known for decades if not centuries.

In math, it is even easier to rediscover what has already been discovered. For example, in A000045, amidst the flurry of comments, we find:

• The ratios F(n+1)/F(n) for n>0 are the convergents to the simple continued fraction expansion of the golden section. - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Dec 19 2004
• F(n+1)/F(n) is also the Farey fraction sequence (see A097545 for explanation) for the golden ratio, which is the only number whose Farey fractions and continued fractions are the same. - Joshua Zucker (joshua.zucker(AT)stanfordalumni.org), May 08 2006
• For n>=1, F(n)=round(log_2(2^{\phi*F(n-1)}+2^{\phi*F(n-2)})), where \phi is Golden ratio. [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Jun 24 2010, Jun 27 2010]
• a(n+1)=ceil(phi*a(n)), if n is even and a(n+1)=floor(phi*a(n)), if n is odd (phi=golden ratio). [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Jul 01 2010]

I am not saying that there is nothing left to discover about the Fibonacci numbers. Even though Thomas Koshy has written an entire book about Fibonacci and Lucas numbers (see if your library has it, by the way) there are still things we don't know and which may yet be discovered. Alonso del Arte 18:30, 14 January 2012 (UTC)

"Golden sequence"

I wanted to make a comment on the essay in your user page.

You write

And since you will only come closer (because of the limit), this proves that you will always be able to round to the next number.

This is incorrect; as an example, the sequence $a(3)=2,a(2n)=F_{2n},a(2n+1)=F_{2n+1}+10^{100}{\text{ for }}n>1$ has $\lim _{n\to \infty }{\frac {a(n)}{a(n-1)}}=\phi$ and $a(3)=\operatorname {round} (\phi \cdot a(2))$ but $a(4)\neq \operatorname {round} (\phi \cdot a(3)).$

In particular it is not true that because a(n)/a(n+1) becomes arbitrarily close to L that a(n) - L a(n-1) becomes arbitrarily close to 0. In fact, the latter is rare!

It happens that $F_{n+1}=\operatorname {round} (\phi \cdot F_{n})$ is true for n > 1, but the asymptotics aren't enough to prove it.

Charles R Greathouse IV 23:15, 1 February 2012 (UTC)

Last digit of Fibonacci numbers

I see you're now interested in Fibonacci numbers mod 10. Are you perhaps familiar with A001175?

I followed the link to your paper, but unfortunately it was unreadable:

Does it display differently for you, I wonder? [1] I notice that only your paper seems to be affected; I could read the others without difficulty.

Charles R Greathouse IV 04:45, 17 September 2013 (UTC)