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let $m$ is positive numbers,and such $m\ge 5$,and $$A_{m+1}=\overline{1234\cdots m}=1\times (m+1)^{m-1}+2\times (m+1)^{m-2}+\cdots+(m-1)\times (m+1)+m$$(or see (

and positive integer number $a$ such $$\gcd{(a,m)}=1,\left[\dfrac{a}{m}\right]=\left[\dfrac{a}{m+1}\right]$$

show that: $$aA_{m+1}=\overline{\sigma_{0}\sigma_{1}\sigma_{2}\cdots\sigma_{m}}$$ where a permutation $\sigma_{0},\sigma_{1},\sigma_{2},\cdots,\sigma_{m}$ of $0,1,2,3,\cdots,m$ is alternating.and $[x]$ is the largest integer not greater than x.

My idea: since $$\left[\dfrac{a}{m}\right]=\left[\dfrac{a}{m+1}\right]\le\dfrac{a}{m+1}$$ then we have $$\left[\dfrac{a}{m}\right]\le a-m\left[\dfrac{a}{m}\right]$$ let $a=pm+r,p,r\in N,0\le r<m$,then we have $$p\le r<m$$. then I can't. I fell this reslut is interesting, I don't know somepaper have this reserch?Thank you for you help

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I don't know what you mean by the notation, $\overline{1234\cdots m}$. – Gerry Myerson Mar 12 '14 at 12:30
@GerryMyerson´╝îhello,can see – math110 Mar 12 '14 at 12:37
If $m=9,a=70$, we have $\lfloor70/10\rfloor=\lfloor70/9\rfloor$ and $A_{10}\times70=123456789\times70=8641975230$, which is a permutation but it is not alternating. Am I missing something? – chubakueno Mar 12 '14 at 19:46
So, the notation seems to mean, 1234...m, interpreted to base $m+1$. Why not just say that? And what is the $n$ in $A_{n+1}$? It doesn't seem to appear anywhere else in the question. – Gerry Myerson Mar 12 '14 at 21:59

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