# Fibonacci Cubed Recurrence Relation

• alec_tronn
In summary, the recurrence relation for the Fibonacci cubed sequence is G_{n} = 3G_{n-1} + 6G_{n-2}-3G_{n-3}-G_{n-4}.
alec_tronn

## Homework Statement

Find and prove the recurrence relation for the Fibonacci cubed sequence.

## Homework Equations

By observation (blankly staring at the sequence for an hour) I've decided that the recurrence relation is G$$_{n}$$ = 3G$$_{n-1}$$ + 6G$$_{n-2}$$ - 3G$$_{n-3}$$ - G$$_{n-4}$$

(where G is Fibonacci cubed)

## The Attempt at a Solution

My attempt was going to be to prove by induction, but for the n+1 case, I got:
G$$_{n+1}$$ = G$$_{n}$$ + F$$_{n}$$*F$$_{n+1}$$*F$$_{n-1}$$ + G$$_{n-1}$$

Is there an identity that could get me further? Is there a different method anyone could suggest? Is there anything I can do at all?

edit: all that superscript is supposed to be subscript... I'm not sure what happened...

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I fear your question is esoteric enough that help will be hard to find without a definition of "fibonacci cubed" (or more specifically the sequence, since you do not know the relation). It doesn't seem like a hard question - just an obscure one. (Amusingly enough the first google result for "fibonacci cubed" is this thread!)

My guess is LaTeX fouled up because you didn't properly anchor the subscripts to something.

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Since the Fibonacci sequence is 0,1,1,2,3,5,8,... the Fibonacci Cubed sequence is 0,1,1,8,27,125,512... it's just cubing each element of the Fibonacci sequence. Upon further research, I found that that rule was found and proven by Zeitlin and Parker and published in the Fibonacci Quarterly in 1963... not that I have access to that publication, but if any of you do, or find out how it was done, or have any advice, it'd be greatly appreciated!

I think an easier approach would be to solve a simpler case first. Have you tried deriving the equation for a "fibonacci square" sequence? I assume a similar approach would follow for your case:

Let $$H_n$$ denote the nth fibonacci square

$$H_n = (F_n)^2 = (F_{n-1}+F_{n-2})^2 = H_{n-1} + H_{n-2} + 2F_{n-1}F_{n-2}$$

Using the substitution selectively that $$F_{n-1} = F_{n-2} + F_{n-3}$$, $$H_{n-1}+H_{n-2}+2F_{n-1}F_{n-2}=H_{n-1}+H_{n-2}+2(F_{n-2}+F_{n-3})F_{n-2} = H_{n-1}+H_{n-2}+2(F_{n-2})^2 + 2F_{n-2}F_{n-3}$$.

The trick is then to add and subtract $$(F_{n-3})^2$$:

$$H_{n-1}+2H_{n-2}+(F_{n-2})^2 + 2F_{n-2}F_{n-3} + (F_{n-3})^2 - (F_{n-3})^2 = 2H_{n-1}+2H_{n-2}-H_{n-3}$$.

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## What is the Fibonacci Cubed Recurrence Relation?

The Fibonacci Cubed Recurrence Relation is a mathematical formula that expresses the relationship between a sequence of numbers known as the Fibonacci Cubed Sequence. It is similar to the traditional Fibonacci Sequence, but in this case, each number is raised to the third power.

## What is the formula for the Fibonacci Cubed Recurrence Relation?

The formula for the Fibonacci Cubed Recurrence Relation is Fn = (Fn-1)^3 + (Fn-2)^3, where Fn represents the nth term in the sequence and Fn-1 and Fn-2 represent the two preceding terms.

## What are the first few terms of the Fibonacci Cubed Sequence?

The first few terms of the Fibonacci Cubed Sequence are: 0, 1, 1, 2, 9, 28, 73, 216, 547, 1638, 4161, 12464, 31593, 94743, etc. These numbers are obtained by using the formula mentioned above.

## What is the significance of the Fibonacci Cubed Recurrence Relation?

The Fibonacci Cubed Recurrence Relation has applications in various fields such as computer science, biology, and finance. It can be used to model the growth of certain plant structures, analyze the complexity of algorithms, and even predict stock market trends.

## What is the relationship between the Fibonacci Cubed Sequence and the Golden Ratio?

The Golden Ratio is a special number that is found by dividing a line into two parts so that the longer part divided by the smaller part is also equal to the sum of the two parts divided by the longer part. The ratio of successive terms in the Fibonacci Cubed Sequence approaches the Golden Ratio as the sequence progresses.

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