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Let $a_n$ denote last nonzero digit of the factorial $n!$. Is the sequence $a_n$ eventually periodic, that is:

$$(\exists n \exists k \forall m > n )(a_{m+k} = a_m)?$$

The conjecture comes from the fact that $n! = (n-1)! \cdot n$ and similar recurrence relation may be true for $a_n$, because reduction modulo $10$ is a homomorphism of rings $\mathbb Z \to \mathbb Z / 10$. The more I think about this the less likely to true it seems, as multiplication by $5$ turns any (last) even digit into zero.

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closed as off-topic by Namaste, Xam, Siong Thye Goh, Dando18, Henrik Aug 11 '17 at 22:18

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It is never periodic. Here is the listing for this sequence in the OEIS.

"This sequence is not ultimately periodic. This can be deduced from the fact that the sequence can be obtained as a fixed point of a morphism."

https://oeis.org/A008904

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  • $\begingroup$ And can you explain what that statement means? Morphism of what???? $\endgroup$ – Rob Arthan Aug 11 '17 at 21:55
  • $\begingroup$ As a fixed point of what morphism? Between what kind of structures? Spaces of sequences? $\endgroup$ – R. Suwalski Aug 11 '17 at 21:57
  • $\begingroup$ thinking out loud if $\mod_{10} {(m+k)!} = m!\prod_{i=1}^{k}{\left(m+i\right)}$ and if (I should check this before posting) $\mod{ab} = \left(\mod{a}\right) \left(\mod{b}\right)$ then $\mod_{10}{\left(\prod_{i=1}^{k}{\left(m+i\right)}\right)} = 1$ always? my apologies if my modulo arithmetic is seriously flawed $\endgroup$ – phdmba7of12 Aug 11 '17 at 22:26

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