I've seen on this site questions asking about rules which would generate a sequence which deviates from say, $2n$ and generate different sequence up to infinity.

Now, I know this is possible.

What I am asking is actually a general question, not only regarding this specific sequence of $2n$, but it would be good enough if you answered even only this specific sequence of $2n$.

Suppose I give you a sequence like $2, 4, 6, 8$. Can you create a formula apart and different from the obvious $2n$ which gets $n$ and delivers $A(n)$, (Which is known to be possible), but while keeping the sequence equidistributed?

  • $\begingroup$ This is not clear. What does it mean for a sequence to be equidistributed? $\endgroup$
    – lulu
    May 10, 2018 at 22:22
  • $\begingroup$ en.wikipedia.org/wiki/Equidistributed_sequence $\endgroup$
    – bilanush
    May 10, 2018 at 22:23
  • $\begingroup$ That notion refers to a collection of points in an interval. How is $\{2n\}$ equidistributed? $\endgroup$
    – lulu
    May 10, 2018 at 22:24
  • $\begingroup$ Do the limit thing to it and you get 1/2 $\endgroup$
    – bilanush
    May 10, 2018 at 22:25
  • 2
    $\begingroup$ That isn't any sort of equidistribution. But if that's all you want, just take $A_n=2n$ for $n\neq 5$ but $a_5=-17$ or any other number you like. $\endgroup$
    – lulu
    May 10, 2018 at 22:33

1 Answer 1


I guess what you want is a sequence that

  1. starts with $\{2, 4, 6, 8,\ldots\}$ but is not the set of even numbers, which is described by $A_n = 2n$.
  2. has a density $d(A_n)$ that that is "a measure of what part of the sequence of all natural numbers belongs to a given sequence" by satisfying $$\liminf_{m \to \infty} \frac{ \text{number of terms}~A_n \leq m}m = c ~, \quad \text{where $c$ is a constant.}$$

Actually the two examples given by @lulu are legitimate sequences with the above properties.

If you want something more interesting:

$A_n \equiv \lfloor n\sqrt{5}\rfloor \quad \text{for}~n = 1,2,3,4,\ldots \quad \text{namely:}~\lfloor \sqrt{5}\rfloor,\, \lfloor 2\sqrt{5}\rfloor,\, \lfloor 3\sqrt{5}\rfloor,\, \lfloor 4\sqrt{5}\rfloor,\ldots$

where the $\lfloor \sqrt{5}\rfloor = 2$ etc is the flooring to the immediate smaller integer.

This is the Beaty Sequence of $\sqrt{5}$, also documented in A022839 in the Online Encyclopedia of Integer Sequences (abbreviated $\color{magenta}{OEIS}$). The first few terms are (showing forty-two terms, colored every ten)

\begin{align*} A_n &= 2, 4, 6, 8, 11, 13, 15, 17, 20, 22, \color{magenta}{24, 26, 29, 31, 33, 35, 38, 40, 42, 44} \\ &\hspace{48pt},46, 49, 51, 53, 55, 58, 60, 62, 64, 67, \color{blue}{69, 71, 73, 76, 78, 80, 82, 84, 87, 89}, 91, 93\ldots\end{align*}

The asymptotic density of a sequence in general is difficult to obtain, but it is easy in this case.

Roughly speaking, the number of terms increases every $\sqrt{5}$. I guess this is what you want, the seqeunce being (misnomer) equidistributed. Upon taking the limit on $m$ (one can also reformulate it to take the limit on $n$), we have $$d(A_n)=\liminf_{m \to \infty} \frac{\text{# of}~ A_n \leq m}m = 1/\sqrt{5}$$ as desired. This is because the flooring makes no difference when it comes to counting the number of terms, except possibly allowing one additional term (the largest) being squeezed into the "threshold" $m$. This exception raises the "finite-density" locally, but we are taking the limit-inferior (unachieved lower bound) so it is the intuitive $1/\surd 5$.

I would like to point you to making good use of OEIS. Conduct searches like so and one will get more results than one can handle. Below are some relatively easily-accessible cases:

then there's A080037 (another simple flooring) on the 4th page, then A067946 ($5^n+1\,|n$) and A057195 ($2^n+7 \overset{?}{=}$prime) with other exponentiation related sequences on the 5th page.

Note that the leading $0$ (or $1$) can be removed just by shifting the index (redefining the sequence).


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