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Construct a sequence which has infinitely many limit points

My attempt: \begin{equation} (a_n)_n = (1,1,2,1,2,3,1,2,3,4,1,2,3,4,5\dots) \end{equation} How do I write a proper formula for my sequence?

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combo student
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  • What's wrong with the definition you gave? A sequence doesn't require an explicit formula to be well-defined. – Umberto P. Apr 11 '16 at 20:56
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    You don't really need formulas, your sequence is perfectly well-defined. If you want something with a closed form and uncountable amount of limit points, try $a_n=\cos(n)$ for example. – Gabriel Romon Apr 11 '16 at 20:57
  • If you want one with all the limit points, you can enumerate the rationals. The construction is a lot easier if you don't care about having rationals appearing more than once in the list, using (for example) the Cantor pairing function to list the pairs of naturals and then taking their quotient. – Patrick Stevens Apr 11 '16 at 21:26
  • $$a_n = n - \frac12\left\lceil\frac{\sqrt{8n+1}-1}{2}\right\rceil \left(\left\lceil\frac{\sqrt{8n+1}-1}{2}\right\rceil - 1 \right)$$ – achille hui Apr 11 '16 at 23:50

5 Answers5

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It may not be your exact sequence, but plays the same trick: $a_n=n-\lfloor \sqrt n\rfloor ^2$.

Hagen von Eitzen
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Seems your sequence is this one:
$a_n = 1 + (n-T_m-m)$ where m is the largest integer
such that the triangular number $T_m \le n$

peter.petrov
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This also works: $$ a_n=\chi_{\mathbb N}(n)+\chi_{2\mathbb N}(n)+ \cdots +\chi_{n\mathbb N}(n)=\sum_{k=1}^{n} \chi_{k\mathbb N}(n) $$ that is equal to the number of divisors of $n$.

Marco Disce
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Consider the sequence

$$1,1/2,1,1/3,2/3,1,1/4,2/4,3/4,1,\dots $$

Then every real number in $[0,1]$ is a limit point of this sequence.

zhw.
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Consider one of the enumerations of the positive rstionals.

This has all the positive reals as limits.

marty cohen
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