Fibonacci-LIKE sequence predecessor prediction formula?

[sarossell]

Any help with this would be greatly appreciated!

I am trying to figure out how to PRECISELY determine the predecessor in a Fibonacci-LIKE sequence, not a TRUE Fibonacci sequence. I have attempted to use (1+sqrt(5n^2)-n)/2 but it only works for TRUE Fibonacci numbers. By Fibonacci-LIKE sequence, I mean ANY number that is NOT actually a true Fibonacci number but is in a similar additive sequence. For example, if I have the randomly chosen number "50896034545" and I want to determine what lesser number would precede it in a Fibonacci-LIKE sequence so that I could continually subtract each successive answer from the previous to form a sequence, I could simply divide by phi (1.618...) and then subtract the value from the given number and repeat.

HOWEVER, when I try to predict the successor value from the smallest value in the sequence so that I can repeatedly add the values to get back to my original number, it's never quite accurate enough in NON-TRUE Fibonacci sequences.

IN FACT, the inaccuracy is completely predictable and follows a saw tooth pattern of successively higher and lower value than it should predict which in itself is inaccurate by increasing values of Fibonacci-LIKE numbers.

What I'm asking is if there are more accurate predecessor and successor formulas for Fibonacci-LIKE sequences. That is not to say, more accurate values of phi. I've already tried phi to 1,500 digits and it's still the same problem.

Thank you!

[deleted-2574]

Hi sarossell,

Two sets of questions:

1. What is the purpose of your inquiry? Do you have a science project? This site is designed to help people doing science projects.
2. Assuming you are doing a science project, what are the values of the numbers in the Fibonacci-LIKE sequence?

[deleted-71487]

Well, I can think of two issues that might be making this difficult.

1) the ratio of successive Fibo-like numbers only *approaches* phi as the sequence increases.

You can see this obviously in your example by choosing 50896034544 as your first number in the sequence (1 is traditionally the second number, but of course it can be anything).

Obviously the ratio of your randomly chosen number to the previous number in the sequence (1) is nothing like phi.

In general, I don't think there's any formula that can uniquely backtrack a Fibonacci-like sequence without some pretty severe constraints, and even then, it will inevitably become less accurate the further back you go.

Even with the *actual* Fibonacci sequence, the ratio of successive numbers moves farther from phi as you go back until it's oscillating around quite substantially as you get towards the first few terms. E.g. 1, 2, and 3/2 are the first 3 ratios.

2) The precision of easily available floating point numbers on a computer is a lot lower than the 1500 digits you've mentioned. Have you written or found an infinite precision math package to use for this purpose?

[sarossell]

I use xlPrecision v.2.0 which is an incredibly useful add-in for Microsoft Excel for arbitrary precision mathematics. It formats large numbers as long text and then performs whatever functions you require.

As for whether this is a science project in the sense that a grade school student made be assigned a project, the answer is "No". I am simply a mathematics hobbyist. I had hoped I would not be shunned for this and instead embraced for bringing an unusual problem to light. I do appreciate the kind replies to my inquiry.

[sarossell]

Although it is true that the ratio between successive numbers in the true Fibonacci sequence only approximates phi, it is still possible to precisely calculate the next larger value using the successor formula 1+sqrt(5n^2)+n)/2. However with generalized Fibonacci-like sequences, this does not work. But it is inaccurate in a very precise manner, being either slightly higher or lower than the actual value. I suspect the solution may include a power of negative phi (-phi^n) to allow for the alternative addition or subtraction of a value to maintain accuracy but I'm not sure how to implement it.

In the true Fibonacci sequence (starting with 0, 1, 1, 2...), the above formula will accurately calculate each successively larger number in the sequence - giving the same number you would get from simply adding two consecutive numbers repetitively. This is not true for generalized Fibonacci sequences. If you start a sequence for example with 12, 17, 29, 46..., by the eleventh term, you begin to approximate the ratio phi pretty closely. But the above formula which works so well for true Fibonacci sequences will not accurately calculate the next value properly - NO MATTER HOW LARGE THE NUMBER IS. And it will in fact be inaccurate in a manner that appears to be a function of phi, offering values either slightly higher or lower than expected.

[deleted-2574]

Hi sarossell,

The Fibonacci-like numbers go by the term "Lucas numbers." So, you can put "lucas number" into a search engine to aid your research.

I was not trying to "shun" you, but the primary goal of this bulletin board (and hence its volunteers) is to help students doing science projects.

[deleted-71487]

I'm not particularly an expert on Fibonacci sequences. I was pretty much going by a statement in this web page: http://www.mcs.surrey.ac.uk/Personal/R. ... asNbs.html that says that all sequences formed by adding the previous 2 values eventually converge to having a ratio between successive values of phi.

That site has pretty much everything anyone would ever want to know about the Fibonacci sequence and phi... perhaps buried somewhere in the hundreds of pages on it is an answer to your question.

I'm still pretty convinced that there's no general formula for predicting the predecessor of these kinds of sequence given a sample value unless you constrain the problem heavily (such as saying that the value is far enough into the sequence so that *for that sequence* the ratio between successive terms is close enough to phi (or whatever, if that site is wrong about the convergence) that dividing by it will round to the correct value, but that's kind of assuming the conclusion in a way).

I'm also guessing that any solution you do find will depend on the starting values of the particular sequence, rather than being some kind of universal answer... but that's just an intuition.