Bases Called Into Question

Number Theory Level 4

Consider the sequence defined by $a_n=\smash[b]{1\, \underbrace{01\cdots01\cdots01\,}_{n \, \text{times}}}$ .

That is, $a_n$ is $1$ followed by $01$ $n$ times ( $n \in \mathbb{N}$ ). The sequence's first terms are $a_1=101$ , $a_2=10101$ , $a_3=1010101$ , $a_4=101010101$ , ...

Therefore, $a_n$ is composite in any base $b$ $(b \in \mathbb{N}$ , $b\geq2)$ for:

All values of

n

All values of

n

, except one None of the other answers At least 2 values of

n

, but not infinitely many All prime values of

n

1 solution

Qweros Bistoros
Aug 12, 2020

$a_n = \sum_{k=0}^{n} b^{2k} = \dfrac{b^{2n+2}-1}{b^2-1} = \dfrac{(b^{n+1}+1)(b^{n+1}-1)}{b^2-1}$

If $n>1$ then $b^{n+1}-1 > b^2-1$ and $a_n$ composite

$a_1 = b^2+1$ can be prime f.i. $b=2$

A little more detail is needed here to prove that $a_n$ is composite for $n \ge 2$ .

if $n$ is even then $b-1$ divides $b^{n+1}-1$ and $b+1$ divides $b^{n+1}+1$ (nontrivially each time),
if $n$ is odd then $b^2-1$ divides $b^{n+1}-1$ nontrivially.

Mark Hennings - 9 months, 4 weeks ago

Bases Called Into Question

1 solution

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