nnbonacci Numbers?

Hi Every33\dfrac{3}{3}

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

nnbonacci Sequence:

This sequence starts with nn ones, and then every subsequent number is obtained by adding the previous nn numbers.

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

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

For n=4n=4, lets call it the 44bonacci Sequence.

And so on...

Can someone please prove that the ratio of two consecutive terms in any nnbonacci Sequence approaches to a root of xn=xn−1+xn−2+......+x2+x1+1\displaystyle x^n=x^{n-1}+x^{n-2}+......+x^2+x^1+1

Thanks!

#Calculus #Sequences #Satvik #Nbonacci

Note by Satvik Golechha
6 years, 5 months ago

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1 vote

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Comments

Let's say that xx is the ratio between two consecutive terms, being a constant for very large nbonacci numbers. Then for nn terms added up, the result is

∑k=0n−1xka=xn−1x−1a\displaystyle \sum _{ k=0 }^{ n-1 }{ { x }^{ k }a=\dfrac { { x }^{ n }-1 }{ x-1 } } a

while the last term of this sum is xn−1a{ x }^{ n-1 }a

The ratio of the two is xx, so we solve for xx the following equation

(∑k=0n−1xka)(1xn−1a)=x \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

xn(2−x)=1{ x }^{ n }(2-x)=1

and eventually the equation you have.

Michael Mendrin - 6 years, 5 months ago

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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.

Calvin Lin Staff - 6 years, 5 months ago

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 - 6 years, 5 months ago
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