some work i've been doing on fibonacci sequences

[hermanntrude]

OK so i'm not a big mathemetician but I was reading up on number sequences the other day and of course I found a lot of stuff on fibonacci sequences. I already knew two things:

 

a fibonacci sequence is made by adding the two previous digits to get the third and then repeating

 

any two consecutive numbers in the sequence when divided by each other will produce an approximation to the golden ratio, and the further into the sequence you go the closer the ratio gets to the golden ratio.

 

What I didn't know was this:

 

You can start a fibonacci sequence with any two numbers and it will STILL tend towards the golden ratio.

 

So I wondered whether there was a relationship between which numbers you start with and how quickly it tends towards the golden ratio. So I used excel (badly... i don't know any programming so a more efficient method was out of reach) and plotted a 100 x 100 grid, each square containing a number representing how many numbers were required in the fibonacci sequence starting with the column and row numbers for that square. I stopped the iterations when the ratio reached between 1.6180338 and 1.6180350. I then graphed the plot as a contour plot, with colours representing the lowest numbers to make it clear which numbers make the most efficient golden-ratio-calculating-fibonacci-sequence. Unsurprisingly the best ones were in a straight line representing the pairs of numbers which came closest to representing the golden ratio on their own. However, the line contained certain pairs which were better than others.

 

I intend to draw a larger graph with the help of some programming, but i'm seeing an interesting pattern in the "good" points on the line. Points which take identical numbers of iterations are regularly spaced at first, but then when the numbers get higher, new points are added to the regular spacing. The distance between the regular points divided by the distance between a regular point and a new one is the golden ratio.

 

I'm not sure of the significance of these results or whether this has been investigated before but I thought you guys might want to discuss it. I'll upload better graphs when I get them.

[the tree]

That's a pretty interesting question, good luck with it. In case you want to read around it, the general form of a Fibonacci sequence is a called a Lucas sequence. Also, if you look at the Wikipedia article on the Fibonacci sequence you'll find an inductive proof for the limiting ratio and it'll confirm that it works with any two seed numbers.

[hermanntrude]

thatnks for that tidbit of info, tree... I had found the lucas numbers while reading but not the lucas sequences page. However, no mention is made of the relationship between lucas sequences and the golden ratio.

[timo]

I don't find this equal-spacing of "good" points surprising, at least if one assumes that the convergence is better the closer the initial ratio was.

 

If you hit the point (x,y) then you probably come very close to the point (2x,2y) later (depending on where you start and how you increase) which has the same ratio. For the good points in between: You might hit a point (trunc(1.5x), trunc(1.5y)), 1.5 being an arbitrary number and trunc meaning some kind of rounding to whatever discretization grid you have chosen, which might be even closer to the golden ratio - note that due to the truncation it does not have the same initial ratio as (x,y). The likelihood to come close to (2 * trunc(1.5x), 2* trunc(1.5y)) later in equidistant steps should then also be there. So should the appearance of new even better in-between pairs.

The explanation is a bit wave-handed, admittedly. But at least if you chose the grid for the two numbers equidistant starting from 0 (exluding (0,0), of course) that should make sense.

[hermanntrude]

I don't find this equal-spacing of "good" points surprising, at least if one assumes that the convergence is better the closer the initial ratio was.

 

If you hit the point (x,y) then you probably come very close to the point (2x,2y) later (depending on where you start and how you increase) which has the same ratio. For the good points in between: You might hit a point (trunc(1.5x), trunc(1.5y)), 1.5 being an arbitrary number and trunc meaning some kind of rounding to whatever discretization grid you have chosen, which might be even closer to the golden ratio - note that due to the truncation it does not have the same initial ratio as (x,y). The likelihood to come close to (2 * trunc(1.5x), 2* trunc(1.5y)) later in equidistant steps should then also be there. So should the appearance of new even better in-between pairs.

The explanation is a bit wave-handed, admittedly. But at least if you chose the grid for the two numbers equidistant starting from 0 (exluding (0,0), of course) that should make sense.

 

i'm not using non-integers at all.

 

Interestingly the regular pattern doesn't arise from the repeated meeting of numbers with the same initial ratio. Someone pointed this out to me by suggesting that 55 and 88 are a good pair because 5 and 8 are in the original fibonnacci sequence and 55 and 88 are simply 11 times each of those numbers. BUT, i realised he was wrong to suggest that, and in fact 5 and 8 takes a lot longer than 55 and 88 (55 and 88 is my best data point with only 9 iterations). 8/5 is equal to 88/55, but, 13/8 (the next two numbers in the original fibonacci sequence) is NOT equal to 143/88.

 

I intend to re-examine my data and pick out those points which are the result of numbers which are multiples of fibonacci numbers, to see if they are always "good" points. I suspect, actually that prime-number multiples of fibonacci numbers might be the best ones, although i'm not really sure why.

[timo]

I am somewhat confused now. The factor should cancel in the nominator and the denominator if I am not completely wrong:

x/y -> (x+y)/x

2x/2y -> (2x+2y)/2x = (x+y)/x

 

In fact, 13*11 = 130+13 = 143, hence 143/88 = (13*11)/(8*11) = 13/8. Or should I better go to bed (or find a calculator)?

 

What I said should also (in fact: especially) hold for integers, too.

[hermanntrude]

whoops. You're right. I had better check my numbers, because that suggests I should have a duplicate point with 9 iterations for 5 and 8