Puzzle time - the Fibonacci numbers

[Horace]

It's a computer language. Computer programmers use computer languages to "talk" to computers and tell them what to do!

[jon-nyc]

Interesting observation:

Cycle lengths 2^n = 3*2^(n-1) at least within the first 100. Does that continue?

[Klaus]

@jon-nyc said in Puzzle time - the Fibonacci numbers:

What language is that?

Haskell.

I bet if you do the same thing in your favorite language you need at least twice as much code, and the code will be less extensible. (throws gauntlet)

[jon-nyc]

I use 8086 assembler for such tasks.

[Doctor Phibes]

Are you two about to have a 'my dick is smaller' contest?

[Klaus]

@Doctor-Phibes Did you know that dick sizes have a Fibonacci distribution? It's a distribution with a very long tail...

<insert @George-K rimshot image macro here>

[jon-nyc]

@jon-nyc said in Puzzle time - the Fibonacci numbers:

Interesting observation:

Cycle lengths 2^n = 3*2^(n-1) at least within the first 100. Does that continue?

bump for Klaus and his little 4 line program.

[Klaus]

@jon-nyc said in Puzzle time - the Fibonacci numbers:

@jon-nyc said in Puzzle time - the Fibonacci numbers:

Interesting observation:

Cycle lengths 2^n = 3*2^(n-1) at least within the first 100. Does that continue?

bump for Klaus and his little 4 line program.

Can you explain? You mean that the cycle length of Fib mod (2^n) is the same as Fib mod (3*2^(n-1))? That doesn't seem to be true.

I uploaded the first 10,000 cycle lengths here, if you want to check this yourself.
https://www.heypasteit.com/clip/0IV18W

[jon-nyc]

No, the cycle length of Fib mod (2^n) = 3*2^(n-1))

Check Fmod2, Fmod4, Fmod8, Fmod16 etc for their cycle lengths. You’ll get 3,6,12,24,48 etc. I’m wondering if it holds. I think it will.

[Klaus]

Here's a list of pairs where the first number shows the "n" (but only for powers of 2) and the second one the associated cycle length. Maybe I missunderstood something but your conjecture doesn't seem to hold.

[(2,6),(4,24),(8,60),(16,24),(32,36),(64,120),(128,420),(256,264),(512,516),(1024,72),(2048,600),(4096,1368),(8192,720)]

[jon-nyc]

But your original list is this one:

(left side numbering mine, right side list yours)

[Klaus]

Ah, looks like an "off by two" error somewhere. Hmm....

[Klaus]

Turned out to be two "off by 1" errors that have to do with indices starting at 0 and not 1.

What about this result?

[(2,3),(4,6),(8,12),(16,24),(32,48),(64,96),(128,192),(256,384),(512,768),(1024,1536),(2048,3072),(4096,6144),(8192,12288)]

[Klaus]

It looks like linear relations between the cycle length happen quite frequently. This is a scatter plot of the cycle lengths I computed above (haven't used R in a while - fun for plotting data!). Observe all the dotted lines. The points you identified are on one of those lines. Also interesting to see a few outliers.

I have marked the points you are interested in in red. You see that there are way more points on that line.

[jon-nyc]

Interesting.

So it turns out Fibonacci cycle lengths a thing people study. They’re called Pisano numbers and understanding them is really about understanding Pisano numbers for prime powers.

A neat property is that if n and m are coprime then period(mn) is least common multiple of period(n) and period(m).

[jon-nyc]

Wait a minute, is that really your graph up there? This is from Wiki

[Klaus]

Yes, it's my graph. I haven't seen the WIki graph yet, but it shows the same data, so of course it's similar. It's an interesting coincidence that they also cut off at 10,000.

[Klaus]

Just for fun, here's the continuation of the dataset for cycle lengths up to 20000.