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 '18 at 22:22
  • $\begingroup$ en.wikipedia.org/wiki/Equidistributed_sequence $\endgroup$ – bilanush May 10 '18 at 22:23
  • $\begingroup$ That notion refers to a collection of points in an interval. How is $\{2n\}$ equidistributed? $\endgroup$ – lulu May 10 '18 at 22:24
  • $\begingroup$ Do the limit thing to it and you get 1/2 $\endgroup$ – bilanush May 10 '18 at 22:25
  • 1
    $\begingroup$ 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 '18 at 22:33

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 satisfies $\displaystyle\lim_{n \to \infty} \frac{ \text{# of terms}~A_m < n,~~\forall m \leq n }n = c$ 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 (misnomer) equidistributed. Upon taking the limit, the flooring makes no difference, thus

$$\lim_{n \to \infty} d(A_n) = \frac1{ \sqrt{5} }$$

I would like to point you to making use of OEIS. You can just search like so and you'll get results more than you can handle. Below are some elementary or interesting cases from this very search:

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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