Interesting $0, 1$ sequence of numbers,after $n>2, a_n$is composite.

Question

Let us have a finite sequence with only $0$ and $1$ digit in our numbers(it can begin with $0$ too). $a_n$ is the number, which we get if we write our number $n$ times next to each other. Prove, that if $n>2$, our numbers are composite.

It is quite hard to understand, here is some examples. Let us have the $01$ sequence, then:

$a_1=1, a_2=101, a_3=10101...$(the 0 at the beginning doesn't count, you can see that $a_3$ can't be a prime)

If we have the $101$ sequence, then:

$a_1=101, a_2=101101...$

How do I prove this in general? It seems hard, but maybe there is a good tactic? :) Thanks.

The statement is not always true. – Sam Clearman Apr 24 '15 at 00:23 — Sam Clearman, Apr 24 '15 at 00:23

SLeslie · Answer 1 · 2015-04-23T21:15:37.280

2

So for the 101 sequence, this is obvious as $101|a_n$ for all $n$.

The same reasoning will suffice for any sequence beginning with 1 (eg: $10001$, etc.)

If your sequence begins with 0, such as the 01 sequence mentioned, you begin with a truncated sequence due to ignoring the initial zeroes in $a_1$, so the same reasoning won't work.

Let $s=0\ldots1\ldots$ denote your string of length $N$ where the first $n$ digits are 0. let $r=1\ldots$ be the reduced string obtained form deleting the $n$ initial 0's of $s$, so $r$ is of length $N-n$.

If $r \neq 1$, then $a_2 = rs= r0\ldots0r$ is divisible by $r$, and this same idea works inductively for $a_n$.

For $r=1$, then this seems more subtle. I will edit when I figure it out.

edited Apr 23 '15 at 21:15

answered Apr 23 '15 at 20:39

SLeslie

2,742

What if $ r=1$? – Théophile Apr 23 '15 at 20:53
Working through a few examples, there is no congruence satisfied by this final case for the $01$ sequence. More thought needs to be applied to see why these are composite. – SLeslie Apr 23 '15 at 21:04

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

0

If the finite sequence has length $l$, then $a_n$ is divisible by $1 + 10^l + 10^{2l} + \cdots + 10^{(n-1)l}$

So the question reduces to, "is $1 + 10^l + 10^{2l} + \cdots + 10^{(n-1)l}$ composite?"

We have $(1 + 10^l + 10^{2l} + \cdots + 10^{(n-1)l})(10^l - 1) = (1+10^n+10^{2n}+ \cdots +10^{(l-1)n})(10^n - 1)$

Thus if $l>1, n>l$, letting $d = \gcd(10^n - 1, 10^l - 1)$ we have $10^n-1 \over d$ divides $1 + 10^l + 10^{2l} + \cdots + 10^{(n-1)l}$

This shows that $a_n$ is eventually composite for any initial sequence other than $1$.

If the initial sequence is one, the question is, "is $1\cdots11$ composite?"

The answer is, not neccesarily. See http://oeis.org/A004023. This has previously been discussed on math.se at When is the number 11111....1 a prime number?.

edited Apr 13 '17 at 12:20

Community

1

answered Apr 23 '15 at 20:43

Sam Clearman

627

This fails for $a_5$ in the $01$ sequence as $a_5 = 101,010,101$, which is smaller than your answer – SLeslie Apr 23 '15 at 20:59
I was indexing differently, my answer is edited. However my argument does not work for any sequence of the form $0 \cdots 0 1$. – Sam Clearman Apr 23 '15 at 21:02
I see. Yeah, that seems to be the tricky case. In fact, your sequence gives the sequence of this form. – SLeslie Apr 23 '15 at 21:04
These are $(10^{nl}-1)/(10^l-1)$, and the numerator is also divisible by $10^n-1$. So that deals with the case where $l$ is not a multiple of $n$. – Empy2 Apr 23 '15 at 21:17
Does it? What if $l=n$, for example? – Sam Clearman Apr 23 '15 at 21:27

Interesting $0, 1$ sequence of numbers,after $n>2, a_n$is composite.

2 Answers2