Messing around with fibonacci

I’m not any sort of mathmetician, and I’m not very familiar with the ideas about the fibonacci sequence, but I came up with F(n) = (F(n/2) + F((n/2)-3)) F(n/2)

I’m sure someone has come up with this before, so I was wondering if I’ve just rearranged someone else’s definition or what. I didn’t derive this from anything more than the basic F(n) = F(n-1) + F(n-2). I’m also not entirely sure why it works. I started with low numbers like n = 14 and n = 16, but wolframalpha tells me it works with n = 200. Can anyone help me?

Can you elaborate a bit? What happens when n is odd? And how is this supposed to work? Can you show a calculation?

I can’t get this to work for the few examples with even indices I tried.

For example, if F(1) = 1, F(2) = 1, then F(5) = 5, and F(10) = 55,

Then F(10/2) = F(5) = 5. And F(10/2 - 3) = F(2) = 1.

According to the formula given, this should be (5+1)*5 = 30, but F(10) = 55 instead.

For the above, if F(1) = 0, F(2) = 1, then F(5) = 3, and F(10) = 34, and things still don’t work.

Incidentally, this looks a fair amount like a known fact about Lucas numbers.

Lucas numbers are like Fibonacci, except the sequence begins 2, 1, 3, 4, 7, 11, …c instead of 0, 1, 1, 2, 3, 5, …

We do know that: F(2n) / F(n) = L(n), and we know L(n) = F(n-1) + F(n+1), leading to:

F(2n) = F(n)*[F(n-1)+F(n+1)]

Thanks, yeah I didnt test it with odd fibonacci numbers. Its probably not very useful except in instances where you can calculate F(n) and F(n-3) and want to know F(2n)

It’d still be helpful if you showed one example.

It does look an awful lot like how Lucas and Fibonacci numbers are related. Even if that’s not entirely new, seeing something old from a new perspective is usually interesting.

To be honest, calculating Fibonacci numbers at all is of limited practical utility, but lots of math people aren’t interested in practical utility to begin with. So, play around and share what you discover. Odds are somebody else thought of it 500 years ago, but it’s still pretty neat to discover even that much. And there’s always the odd chance you actually do have something undiscovered.

This also works:
F(2n) = F(n)*[F(n)+F(n-1)+F(n-2)+F(n-3)]

Or closer to how the OP put it:
F(n) = (F(n/2) + F((n/2)-1) + F((n/2)-2) + F((n/2)-3)) F(n/2)

Maybe the OP just missed those middle terms?

Here are a couple quick examples:

F(6) [8] = (2F(6/2) + F(6/2 -3)) (F(6/2)) = (2F(3) + F(0)) (F(3)) = ((2 * 2) + 0) * 2 = 8

F(10) [55] = (2F(10/2) + F(10/2-3)) (F(10/2)) = (2F(5) + F(2)) * (F(5)) = ((2 * 5) + 1) * (5) = 55

That helps immensely, actually.

You were missing a “2” in your OP, which makes a huge difference.

So, the expression should actually be:

F(2n) = [2*F(n) + F(n-3)]*F(n)

This is absolutely true and actually is what was mentioned above:

F(n-1) + F(n-2) = F(n), so this expression becomes:

F(2n) = F(n)[F(n)+F(n-1)+F(n-2)+F(n-3)]
F(2n) = F(n)
[F(n)+F(n)+F(n-3)]
F(2n) = F(n)[2F(n) + F(n-3)]

which is what you got.

So, yes, this is something that is known about Fibonacci numbers, but it’s still always kind of neat to rediscover something on your own.

Ah, I cant believe I mistyped it in the OP! Thanks

The chances that you can discover anything new about the Fibonacci sequence are pretty low. There is an entire journal devoted to results about it. In fact, I published a paper in 1975 (in another journal) with results related to the Fibonacci sequence. What is your educational status? If you truly want to publicize a result about this, you would have to learn enough to be able to prove the theorem, not just state it from a few examples:

http://www.fq.math.ca/

http://faculty.evansville.edu/ck6/bstud/fnt.html

I know we’ve had problems with self-declared geniuses thinking they proved some new mathematical truth before, but I didn’t get that sense from the OP. I just got the sense that they were experimenting with the Fibonacci definition and were asking a question about whether it was something people have used before or not.

Oh, I agree that this is just dunkleosteus messing around with numbers and noticing something interesting. Regardless of how old he/she is or what his/her educational status is, I think that’s great. I didn’t mean to criticize him/her at all. I just thought that dunkleosteus should know that there’s already a lot of work done on this already, and he/she should learn about it.

I ask about his/her educational status because that’s important to what I would say. If he/she is 6 years old and noticed this, that would be wonderful. dunkleosteus should get some books about similar things and read them. He/she might have a career in math. On the other hand, if he/she is 93 years old, perhaps it’s a little late to start a new career, but it’s still great that he/she has noticed this, and perhaps he/she would like to learn about other things about the Fibonacci sequence and about how mathematical theorems are proved.

Ah sorry for the confusion. I tried to look up some of the theorems regarding the Fibonacci sequence online, but a lot of what I found didn’t make much sense, so I was hoping I could find someone here who would be able to tell me what I had done- I figured I was probably doing some rearrangement of something someone else had already proved. If you’re looking for the right pronoun, I’m male, and as far as education I’m in my third year at university majoring in computer science.

On any given topic, there are always more complicated theorems than there are simple theorems. It’s part of the nature of complexity. There are usually a few simple theorems, too, but it can be hard to find them in amongst all the complicated ones.

Sure, I can explain to you why it works.

First, let’s say a Fibonacci-type sequence is any sequence where each value is the sum of the previous two. It doesn’t have to have the same starting values as the ordinary one; it can have any starting values you like. It just has to then proceed by the usual Fibonacci rule.

Note that, if you had any value m such that m[sup]2[/sup] = m + 1, then the sequence of powers m[sup]0[/sup], m[sup]1[/sup], m[sup]2[/sup], m[sup]3[/sup], etc., would be a Fibonacci-type sequence. Let’s suppose we have such an m for the time being.

The first term would be 1 (which is as good as 1 + 0 * m) and the next term would be m (which is as good as 0 + 1 * m). After that, each term would be the sum of the previous two (so we would get 1 + 1 * m, and then 1 + 2 * m, and then 2 + 3 * m, and then 3 + 5 * m). In general, we would find that m[sup]n[/sup] = F(n - 1) + F(n) * m. So if we also had some operation H such that H(a + b * m) = b, we could recover F(n) as H(m[sup]n[/sup]). Again, let’s suppose we have such an operation for the time being.

By squaring both sides of the identity m[sup]n[/sup] = F(n - 1) + F(n) * m, we find that m[sup]2n[/sup] = (F(n - 1) + F(n) * m)[sup]2[/sup]. Expanding this out (including using the expansion m[sup]2[/sup] = m + 1), and then applying H, we get F(2n) = H(m[sup]2n[/sup]) = 2F(n - 1)F(n) + F(n)[sup]2[/sup].

Furthermore, we can rewrite 2F(n - 1)F(n) + F(n)[sup]2[/sup] as [2F(n - 1) + F(n)]F(n). And (renoting what others have noted above) we can rewrite the [2F(n - 1) + F(n)] as [F(n - 1) + F(n - 1) + F(n)], which in turn equals [F(n - 3) + F(n - 2) + F(n - 1) + F(n)], which equals [F(n - 3) + F(n) + F(n)], which is to say, [F(n - 3) + 2F(n)]. Plugging this back in, we’ve found that F(2n) = [F(n - 3) + 2F(n)]F(n), just as noted in the OP.

But how do we know there’s such a thing as m and H? Well, just like we can arbitrarily invent an i which squares to -1 and maintain familiar rules of arithmetic to get complex numbers, from which “real” and “imaginary” parts can be extracted, similarly, we can arbitrarily invent an m such that m[sup]2[/sup] = m + 1 and maintain familiar rules of arithmetic to get an appropriate number system for this task (with H being the appropriate part-extracter).

If you’re unhappy doing things so abstractly, you might also think of m as the linear operator on Fibonacci-type sequences which shifts the starting point over by one. The rules of Fibonacci-type sequences are such that m[sup]2[/sup] = m + 1 by definition. We can then think of H(an operator) as the first value of the sequence you get from applying that operator to the Fibonacci-type sequence starting 0, 1, … (i.e., to the standard Fibonacci-type sequence).

If any m works, then why can’t we simply solve for m in the original equation? As expected, one solution is the familiar golden ratio of 1.618. It’s easy to verify that increasing powers indeed follow a Fibonacci-like sequence: 1.0, 1.618, 2.618, 4.236, etc.

Certainly one can invent an m just as was done with i, but if we’re keeping things concrete it seems that it’s easy just to assume m is the golden ratio.

Any m will work so long as you can also define an H to extract b from a + b * m. I glossed over it, but in doing this, we must have in mind that a and b range over some more restrictive account of numbers than the general ones containing m (for otherwise, we would have that, for example, m + 0 * m and 0 + 1 * m were both representations of m, so H(m) would not know whether to extract 0 or 1).

The easiest thing is to say that a and b range over whatever “ordinary” numbers are, and m is, by fiat, some new, extra-ordinary number. Then H is well-defined automatically.

Alternatively, you could say that m is the Golden Ratio (the positive solution to x[sup]2[/sup] = x + 1 obtained by continuity), and restrict a and b to, say, the integers. But then, to know H is well-defined, you have to establish that the Golden Ratio is irrational. Which is not super-difficult, but not fundamentally relevant to the task at hand either.

[For example, we might wish to apply analogous reasoning to the sequence 0, 1, 1, 3, 5, 11, 21, 43, …, where each item is the previous item plus twice the item before that. And we can! But we will need to make use of a root of x[sup]2[/sup] = x + 2, and here it really is important that we allow ourselves to make a new one rather than go hunting in the traditional fashion (for both the traditional roots will be rational, and thus run afoul of the aforementioned issue).]

All right, yeah. Just like we can build operators that extract one coefficient of a polynomial, or infinitesimal, or indeed the imaginary part of a complex number. a+bx=c+dx implies a=c and b=d only if m is an “extra-ordinary” number of some kind. Irrational vs. integer works, but we can simply define m that way by fiat if we want.

If you like, we can also work with pairs of numbers, with all arithmetic done component-wise. Then we can take m to be the pair whose first component is the Golden Ratio 1.618…, and whose second component is the Golden Ratio’s conjugate -0.618… . Ordinary numbers will be pairs whose two components are equal, while this m is clearly extra-ordinary.