Surfing the internet I bumped into a very interesting problem, which I tried to solve, but got no results.

The problem is following: let $h_n$ be the most right non-zero digit of $n!$, for example, $10!=3628800,$ so $h_{10}=8$. The task is to prove that decimal fraction $0,h_1h_2\ldots h_n\ldots$ is irrational.

  • $\begingroup$ You may be able to get a clear answer if you deal with $p$-ary expansion instead, even when $p=2$, binary expansion. $\endgroup$ – Lubin May 31 '16 at 19:19
  • $\begingroup$ Irrationality-proof of this kind are extremely hard, but the continued fraction that I calculated with PARI/GP considering the first 500 digits indicates at least a very large denominator, if the number should actually be rational. If we cannot prove a period in the sequence by exploiting properties of the last non-zero digit for a factorial, we probably cannot do better than calculate as many digits as possible. $\endgroup$ – Peter May 31 '16 at 19:53
  • $\begingroup$ With the factorials upto $10,000!$, I get $19,409$ terms of the continued fraction with maximum $129,375$ (The last few terms may be incorrect). This strongly indicates (but of course in no way proves!) that the number is irrational. $\endgroup$ – Peter May 31 '16 at 19:57

To show that $0,h_1h_2\ldots$ is irrational, it's enought to show that the sequence $h_n$ is not eventually periodic. Assume on contrary that $h_n$ is eventually periodic. Then $h_n$ is eventually periodic as a sequence modulo $5$.

Let $\mathbb Q^+$ denote the multiplicative group of positive rationals. The set of prime numbers constitutes a base of $\mathbb Q^+$ regarded as $\mathbb Z$-module. Consequently, there exists a unique multiplicative group homomorphism $\lambda:\mathbb Q^+\to\mathbb Q^+$ such that: $$\lambda(p)=\begin{cases}p&p\neq 5\\\frac 12&p=5\end{cases}$$ ($p$ prime number). Then we have $h_n\equiv \lambda(n!)\pmod{10}$ for each $n>0$, hence $$h_n\equiv\lambda(n!)\equiv \prod_{k=1}^n\lambda(k)\pmod{5}$$ Note that $\lambda(k)=k$ if $5\nmid k$.

Lemma. Let $u_n(n\in\mathbb N)$ be a sequence in a multiplicative group $G$. If $\prod_k u_k$ is eventually periodic, then $u_n$ is eventually periodic.

For if $$\prod_{k=0}^{n+T}u_k=\prod_{k=0}^{n}u_k$$ for each $n\geq N$, then $$\prod_{k=n+1}^{n+T}u_k=1$$ for each $n\geq N$, hence $$\prod_{k=n+1}^{n+T}u_k=1=\prod_{k=n+2}^{n+1+T}u_k$$ from which $u_{n+1}=u_{n+1+T}$ for each $n\geq N$.

Thus if $h_n$ is eventually periodic, then also $\lambda(k)$ is eventually periodic in the group of units of $\mathbb Z\diagup 5\mathbb Z$, that's $\lambda(n+T)\equiv\lambda(n)\pmod 5$ for each $n\geq N$. Write $T=5^vT'$ with $5\nmid T'$; for $n\geq N$ and $n>v$ we have $\lambda(5^n)\equiv 3^n\pmod 5$ and $$\lambda(5^n+T)=\lambda(5^n+5^vT')=3^v\lambda(5^{n-v}+T')\equiv 3^v(5^{n-v}+T')\equiv 3^vT'\pmod 5$$ thus $T'\equiv 3^{n-v}\pmod 5$ for each $n$ large enought - a contradiction which concludes the proof.


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