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Let L(x) be the equation that gives us the number of possible arrangements of x Legos—L(2)=24, L(3)=1560, L(6)=915,103,765, etc..

I think that this might be true: $$\lim_{x\to \infty} \frac{L\left(x+1\right)}{L\left(x\right)} \approx 100 $$

Can someone prove this to be true or false?

diligar
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    If they could (prove it true or false), then it would be worthy of writing a paper on. A paper linked to in your link discusses the problem and gives upper and lower bound estimates on the rate of growth being between $78.32$ and $191.35$. – JMoravitz Dec 22 '15 at 02:07
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    It is worth reading a more recent paper found here which goes further and generalizes the problem to bricks of arbitrary length/width. One of the authors was cited in the previously linked article while the other author was also an author on the previously linked article. There, they make the claim that a more accurate estimate would be 117. Also of interest is the oeis entry for this problem where they now have nine entries of the sequence calculated. – JMoravitz Dec 22 '15 at 02:23

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I am not sure what is meant by "$\approx$" in the question but the status of this question is:

  1. I don't think there is any way to disprove that $\lim_{n\to\infty}\frac{L(n+1)}{L(n)}=100$ exactly (although I find that highly unlikely).
  2. It is unknown if $\lim_{n\to\infty}\frac{L(n+1)}{L(n)}$ exists.
  3. It is known that $\lim_{n\to\infty}\sqrt[n]{L(n)}$ exists and is in the interval $[78,192]$.
  4. If $\lim_{n\to\infty}\frac{L(n+1)}{L(n)}$ exists, it must hence also be in this interval.
Søren Eilers
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  • It was quite a while ago that I asked this, but I think I was just trying to find a pattern to $L(n+1) / L(n)$, and it seemed that it was closing in on 100---but that was just a guess – diligar Jul 20 '18 at 16:52