Hi Every\(\dfrac{3}{3}\)

I just found something interesting, and wanted to share it. Also, please help me prove it.

**\(n\)bonacci Sequence:**

This sequence starts with \(n\) ones, and then every subsequent number is obtained by adding the previous \(n\) numbers.

For \(n=2\), it's called the Fibonacci Sequence.

For \(n=3\), it's called the Tribonacci Sequence.

For \(n=4\), lets call it the \(4\)bonacci Sequence.

And so on...

Can someone please prove that the ratio of two consecutive terms in any \(n\)bonacci Sequence approaches to a root of \(\displaystyle x^n=x^{n-1}+x^{n-2}+......+x^2+x^1+1\)

Thanks!

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TopNewestLet's say that \(x\) is the ratio between two consecutive terms, being a constant for very large nbonacci numbers. Then for \(n\) terms added up, the result is

\(\displaystyle \sum _{ k=0 }^{ n-1 }{ { x }^{ k }a=\dfrac { { x }^{ n }-1 }{ x-1 } } a\)

while the last term of this sum is \({ x }^{ n-1 }a\)

The ratio of the two is \(x\), so we solve for \(x\) the following equation

\( \left(\displaystyle \sum _{ k=0 }^{ n-1 }{ { x }^{ k }a } \right) \left( \dfrac { 1 }{ { x }^{ n-1 }a } \right) =x\)

A little bit of work gets you

\({ x }^{ n }(2-x)=1\)

and eventually the equation you have. – Michael Mendrin · 2 years, 2 months ago

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– Calvin Lin Staff · 2 years, 2 months ago

A slight concern would be the initial assumption that the ratio does approach a finite number. It is clear that the ratio is bounded between 1 and n, but it is not immediately clear that the ratio must converge to a limit.Log in to reply

For linear recurrence with distinct real roots, this is an immediate result of the characteristic equations of linear recurrence relations.

Care has to be taken when the roots are complex, are when we have distinct roots, as you study the limiting behavior. – Calvin Lin Staff · 2 years, 2 months ago

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