# Is it possible to have a closed formula for a sequence 2, 4, 6, 8, ... and the sequence would be equidistributed?

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?

• This is not clear. What does it mean for a sequence to be equidistributed?
– lulu
May 10, 2018 at 22:22
• en.wikipedia.org/wiki/Equidistributed_sequence May 10, 2018 at 22:23
• That notion refers to a collection of points in an interval. How is $\{2n\}$ equidistributed?
– lulu
May 10, 2018 at 22:24
• Do the limit thing to it and you get 1/2 May 10, 2018 at 22:25
• 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.
– lulu
May 10, 2018 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 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).