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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?

Lee David Chung Lin
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bilanush
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  • This is not clear. What does it mean for a sequence to be equidistributed? – lulu May 10 '18 at 22:22
  • https://en.wikipedia.org/wiki/Equidistributed_sequence – bilanush May 10 '18 at 22:23
  • That notion refers to a collection of points in an interval. How is ${2n}$ equidistributed? – lulu May 10 '18 at 22:24
  • Do the limit thing to it and you get 1/2 – bilanush May 10 '18 at 22:25
  • No idea what you are talking about. Sorry. – lulu May 10 '18 at 22:26
  • You know there is some "density" of certain sequences like say, even the prime numbers have same characteristic of a consistency when you calculate the phi which is all the numbers bellow a certain prim. You do a similar notion of limit to the sequence you would get for 2n you get 0.5. Would you get a lim which approaches a constant had you deviated from 2n? – bilanush May 10 '18 at 22:29
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    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 '18 at 22:33
  • I dont understand. Firstly, how do you know you have a close formula polinomic or even not which describes this. Secondly how would that be with equal density. – bilanush May 10 '18 at 22:38
  • Once again, there is no notion of "equal density" involved here. The limit you sketch is obviously the same for my sequence as for yours (it only differs in one value), and I gave a closed definition of the sequence. Another would be $a_n=2n$ for $n≤4$ and $a_n=2n-1$ for $n>4$. But, honestly, I don't understand what you are looking for, so I think I should stop trying to guess. – lulu May 10 '18 at 22:40
  • Thx for your time. Sorry. I don't speak english well I just know the terms in my language but not in english. I am just looking for some sort of a diffrence between the continuation of 2n to a diffrent formula starting with 2,4,6,8 . Presumably something in the difference should be somehow related to the distribution / density or whatever which I am aware to be pretty unfamiliar with the terms or concepts. Thx again. – bilanush May 10 '18 at 22:46

1 Answers1

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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).

Lee David Chung Lin
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