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Let $u_n$ be a sequence defined on natural numbers (the first term is $u_0$) and the terms are natural numbers ($u_n\in \mathbb{N}$ )

We defined the following sequences:

$$\displaystyle \large x_n=u_{u_n}$$ $$\displaystyle \large y_n=u_{u_n}+1$$

If we know that both $y_n$ and $x_n $ are arithmetic sequences ,how we can prove that $u_n $ is also arithmetic sequence

Iuli
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Brab
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  • http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1561573&sid=b1e33a1bfa4435c6c5b35b24da884c66#p1561573 – only Aug 28 '12 at 08:17
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    As written, the fact that ${y_n}$ is arithmetic is redundant given that ${x_n}$ is. Did you mean: $\displaystyle \large x_n=u_{u_n}$ AND either $\displaystyle \large y_n=x_{u_n}+1$ or $\displaystyle \large y_n=u_{x_n}+1$? – Marconius Jul 09 '15 at 19:51

1 Answers1

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I don't think that $u_n$ must be an arithmetic sequence.

$u_n=a, a+\Delta u, a+2\Delta u, ... , a+(k-1)\Delta u, ...$

$$\Rightarrow x_n=u_{u_n}=a+(u_n-1)\Delta u$$ $$\Rightarrow y_n=u_{u_n}=a+(u_n-1)\Delta u+1$$

In arithmetic series, consecutive terms must have a constant difference. Looking at at $x_n$,

$$a+(u_n-1)\Delta u-[a+(u_{n-1}-1)\Delta u]=\Delta x$$ $$(u_n-u_{n-1})\Delta u=\Delta x$$

$$\Rightarrow u_n-u_{n-1}=\frac{\Delta x}{\Delta u}\neq\Delta u$$ unless $\Delta x=\Delta u^2$, which isn't always the case.

I may have misunderstood the problem, though.

Lucien Jaccon
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