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Given a bijection $g:\mathbb{N}\to\mathbb{N}$, consider the sequence $g(1), g(2), g(3),\ldots.\ $ Does there exist a bijection $f:\mathbb{N}\to\mathbb{N}$ such that $\vert f(n)-f(n+1) \vert = g(n)\quad \forall\ n\in\mathbb{N}?$

Answers to this question show that the answer is affirmative when $g$ is the identity function. But I wonder if it is true in general.

I suspect there is some counter-example, but since I haven't been able to construct one, this suspicion isn't very strong.

Adam Rubinson
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  • Just to avoid misunderstandings: Is $0 \in \Bbb N$ or not? – Martin R Jan 30 '25 at 13:00
  • @MartinR No, $0$ is not in $\mathbb{N}.$ – Adam Rubinson Jan 30 '25 at 13:01
  • FYI My approach can also be extended to bijections $ g: \mathbb{N} \rightarrow \mathbb{Z}^x $ or $ g: \mathbb{Z} \rightarrow \mathbb{Z}^x$. The idea is to have a large $g(2)$ that forces us away, and then selecting values to ensure that $f$ can never come back near $f(1)$ again. – Calvin Lin Feb 14 '25 at 19:53

1 Answers1

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Preamble: We will construct a function $g$ such that there is no bijection $f$ that satisfies $|f(n+1) - f(n) | = g(n)$. The idea is:

  1. If $f(n) < g(n)$, then we must have $f(n+1) = f(n) + g(n)$, so we can force $f(n)$ to increase significantly.
  2. For $g(n)$ to be a bijection, it needs to cover the small values.
  3. If $g(n)$ is alternating large and small, then perhaps we can force $f(n)$ to grow quickly, and hence cannot cover the small values.

As a numerical example,
If $f(1) = 10$ and $g(1) = 1,000$, then we must have $f(2) = f(1) + g(1) = 1010$.
If $g(2) < 10$, then $f(3) \leq f(2) + g(2) = 1020$.
If $g(3) > 1020$, then we must have $f(4) = f(3) + g(3) > 1020$.
If we can show that for $n > 4$, $f(n) > 1020$, then $f$ will not be a bijection.

The linked question avoids this structure because the values of $g(n)$ are close to each other.

Solution: We essentially fill in the details of the idea / numerical example and verify the inequalities that we need.

Define $g(n)$ in the following way:

  • For $ n = 2k-1$, $g(n) = 4^k$
  • For $n = 2k $, $g(n)$ is the smallest positive integer that hasn't been attained by $g(i), i < n$.
    • This ensures $g$ is a bijection.
    • Trivially $g(n) \leq n$ for these values. That is all that we need.

Let $f$ be any function (not necessarily a bijection) such that $|f(n+1) - f(n) | = g(n)$.

Observation: For $n = 2k-1$ and $k \geq 2$, $ \sum_{i < n} g(i) < \frac{2}{5} g(n)$.
Proof: We split the sum for the odd and even terms.
$\sum_{i =1}^{k-1} g(2i-1) \leq \frac{4^{k-1} } { 1 - \frac{1}{4}} = \frac{ 4^k } { 5} $ as the sum of a geometric progression.
$ \sum_{i =1 }^k g(2i) \leq 2 \sum i = \frac{k(k+1) } { 2} < \frac{4^k}{5} $ as an arithmetic progression sum.
Thus $\sum_{i < n} g(i) < 4^k \times \frac{2}{5} $.

Observation: No matter what value $f(1)$ takes, there exists a $K$ such that for all $k > K$, $f(2k) - f(2k-1) = g(2k-1)$.
Proof: Take $ K = \lfloor \log_{4} 5 f(1) \rfloor +2 \geq 2$, so $5f(1) < 4^K$.
For any $k \geq K$, $f(2k-1) \leq f(1) + \sum_{i < 2k-1} g(i) < \frac{1}{5} 4^K + \frac{ 2}{5} 4^k < 4^k = g(2k-1) $.
Hence $f(2k) - f(2k-1) = g(2k-1)$.

Corollary: For $k > K$, $f(2k) > g(2k-1) \geq 4^k > 4^K$.

Corollary: For $n = 2k+1$ and $k > K$, $$f(2k+1) \geq f(2k) - g(2k) \geq f(2k) - 2k > 4^{k} - 2k \geq 4^K.$$

Corollary: Since $f(n) < 4^K $ implies that $n \leq 2K+1 $, hence there aren't enough values for the range of $f$ to contain all of the integers up to $4^K$, thus $f$ is not a bijection.

Calvin Lin
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