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Suppose, some finite digit-sequence is given. Can we prove or disprove, that there is always some number $n$, such that the digit-sequence appears in the decimal-expansion of the number $n!$ ?

If there is no specific pattern in the decimal-expansions of the factorials, this should be the case, but I have no idea how we can check it.

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  • $\begingroup$ Here is a reformulation which may or may not help. It would be the same as to say that if there exists a number $n$ such that $n!$ after division by $10^k$ for some $k$ will be $ = p (\textrm{mod} 10^l)$ where $p$ is any given number (of length $l$). $\endgroup$ – mathreadler Dec 1 '15 at 19:25
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In short: Benford's law is provably true in any base for factorials.

Pick a base $B\ge 2$ and $m\gg 0$ such that $$\tag0B^{m-1}>16.$$ Let $n=\lfloor\log_B(B^m!)\rfloor$ and for $k=0,1,2,\ldots$ let $$\tag1a_k=\frac{(B^m+k)!}{B^{km+n}}.$$ Then $1\le a_0<B$ and $$\tag2\frac{a_k}{a_{k-1}}=1+\frac{k}{B^m}$$ From $(2)$ and $a_{k-1}\ge\ldots\ge a_0\ge 1$ we have $$ a_k- a_{k-1}=a_{k-1}\frac{k}{B^m}\ge \frac{k}{B^m},$$ hence $$a_k\ge 1+B^{-m}\sum_{j=1}^kj=1+\frac{k(k+1)}{2B^m} $$ and specifically (using $(0)$) $$\tag3a_{\frac12B^m}>\frac18B^m >2B$$ Also from $(2)$ we have $$\tag4\lfloor a_{k-1}\rfloor \le \lfloor a_k\rfloor\le \lfloor a_{k-1}\rfloor +1\qquad\text{if }a_{k-1}\le 2B\text{ and }k\le \frac12B^{m-1}.$$ Indeed, this follows from $a_k-a_{k-1}=\frac k{B^m}a_{k-1}\le 1$ under the given bounds for $a_{k-1}$ and $k$.

From $(3)$ and $(4)$ we conclude that $\lfloor a_0\rfloor,\ldots,\lfloor a_{\frac12B^{m-1}}\rfloor$ covers at least the integers $B,\ldots, 2B-1$. Specifically, all possible base-$B$ digits occur as unit digts in some $a_k$, hence as some base-$B$ digit in a factorial.

If we let $B=10^l$, this means that any length-$l$ decimal digit sequence occurs in some factorial.

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  • $\begingroup$ I am sorry, but I cannot follow this proof. Perhaps, you try to explain the proof in words. $\endgroup$ – Peter Dec 1 '15 at 20:42
  • $\begingroup$ @Peter The idea is that - staring from large-power-of-ten factorial - the next factorials are multiplied only by numbers not much larger than a power of ten, hence almost just getting longer and the leading several digits almost unchanged. Then again, the cumulative effect of the many numbers not much larger than that power of ten is so large that the many small changes in the leading digits must accumulate to a "full rotation of the odometer" at least. - Which part of my more formal write up do you have trouble with? I might fill in. $\endgroup$ – Hagen von Eitzen Dec 2 '15 at 20:14
  • $\begingroup$ this was helpful, thank you! $\endgroup$ – Peter Dec 2 '15 at 20:21

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