1
$\begingroup$

$5775$ and $5776$ are two consecutive abundant numbers, i.e., $n$ and $n+1$ such that both $\sigma(n) > 2n$ and $\sigma(n+1) > 2(n+1)$ (see A096399).

Are there any two consecutive numbers $n$ and $(n+1)$ such that both $\sigma(n) > 3n$ and $\sigma(n+1) > 3(n+1)$?

$\endgroup$
0
$\begingroup$

You want numbers such that $\frac{\sigma(n)}{n}>3$.

Recall that if $n=\prod\limits_{i=1}^s p_i^{n_i}$ then $\frac{\sigma(n)}{n}=\prod_{i=1}^s \frac{1-(1/p_i)^{n_i+1}}{1-(1/p)}$.

Now, it is not hard to show that $\prod\limits_{i=1}^\infty\frac{1-(1/p_i)^2}{1-(1/p)}=\prod\limits_{i=1}^\infty 1+\frac{1}{p}$ diverges.

This means that we can find distinct primes $a_1,a_2\dots a_m$ and $b_1,b_2\dots b_l$ so that $\prod\limits_{i=1}^m 1+\frac{1}{a_i}>3$ and $\prod\limits_{i=1}^m 1+\frac{1}{b_i}>3$.

It is a direct consequence of the chinese remainder theorem that there exists and integer $n$ such that $n\equiv 0 \bmod a_i$ and $n\equiv -1 \bmod b_j$ for all suitable $i,j$.

Therefore we must have that $\frac{\sigma(n)}{n}>3$ and $\frac{\sigma(n+1)}{n+1}>3$ as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.