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What is the largest possibly infinite $X$, such that there exist an infinite sequence of integers, $a_1,a_2,a_3...,$ with $a_n=d(1,n)$, with $d(k+1,n)=d(k,n+1)-d(k,n)$ for $k>1$, and where $d(m,n)$ for all integers $X>m>1$ satisfy:
i) $d(m,i)\neq d(m,j)$ for $i\neq j$
ii) $\mathbb{Z}=\{d(m,i)\}_{i>0}$

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Do you have any examples that work even for first and second differences? –  Gerry Myerson Feb 20 '12 at 22:47
I made a mistake in the previous comment, so I had to rewrite it. I apologize for the inconvenience. The solution to the problem as stated is $X=2$, for otherwise we would have $d(3,i)=0$ for some $i$, implying $d(2,i+1)=d(2,i)$, a contradiction. A sequence for $X=2$ is easily constructed as follows: take your favourite sequence $(b_n)_{n=1}^\infty$ that contains each integer exactly once and define $a_n = \sum_{i=1}^{n-1} b_i$. –  Dejan Govc Feb 20 '12 at 23:10

1 Answer 1

up vote 2 down vote accepted

Let $\sigma_0$ be the sequence $\langle 0,1,0,2,0,3,0,4,0,\dots\rangle$. That is, if $\sigma=\langle a_n:n\in\mathbb{Z}^+\rangle$, then $$a_n=\begin{cases} 0,&\text{if }n\text{ is odd}\\ n/2,&\text{if }n\text{ is even}\;. \end{cases}$$

Let $\sigma$ be the sequence obtained from $\sigma_0$ by prepending a $0$ term: $\sigma=\langle 0,0,1,0,2,0,\dots\rangle$. It’s easy to see that the sequence of first differences of $\sigma$ is $\langle 0,1,-1,2,-2,3,-3,\dots\rangle$, whose $n$ term is $-(n-1)/2$ if $n$ is odd and $n/2$ if $n$ is even. Thus, $X=2$ is certainly possible.

However, if the second differences range over $\mathbb{Z}$, one of them must be $0$, and therefore two (consecutive) first differences must be equal, so it’s impossible to have $X\ge 3$.

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