# On the sequence of positive integers satisfying $\sigma(n)\mid (n(\sigma_0(n))^2)$

If my calculations with my computer were rights, when I've consider the sequence of integers $n\geq 1$ satisfying $$\sigma(n)\mid (n(\sigma_0(n))^2),$$ where $\sigma(n)=\sum_{d\mid n}d$ is the sum of divisors function and $\sigma_0(n)=\sum_{d\mid n}1$ the number of such divisors, then such sequence starts as

1, 3, 6, 15, 28, 33, 42, 84, 91...

but I believe that such sequence isn't in The On-Line Encyclopedia of Integer Sequence.

Question. Are right my first terms? Is known such sequence in previous encyclopedia or are there references in the literature?

I've curiosity to know what's about the number of terms in the sequence for the first $10^k$ with $k=4,5,6,\cdots, N$, thats is a plot of the counting function for previous sequence.

Question. Can you provide us a graph of the counting function of such sequence for $10^N$, for a $N$ large? Thanks in advance.

• let $f(n) = n \sigma_0(n)^2$. you have "$\sigma(n) | f(n)$ and $\sigma(m) | f(m)$, $gcd(n,m) = 1$" $\implies$ $\sigma(nm) | f(nm)$. So a question is if there are infinitely many $n$, $gcd(n,3) = 1$, $\sigma(n) | f(n)$ – reuns Sep 13 '16 at 11:40
• Welcome @user1952009 and very thanks much. Then I undertand that you prove a claim about the sequence, but from it is deduced that is a well known sequence? It was surprising to me that the sequence that I've calculated, I believe, that isn't in OEIS. Thanks one more time. – user243301 Sep 13 '16 at 11:41
• part of your sequence is oeis.org/A001599 where you get additional terms by squaring your numerator – Will Jagy Sep 13 '16 at 19:32

The harmonic numbers begin $$1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200, 8128, 8190, 18600, 18620, 27846, 30240, 32760, 55860, 105664,$$ and are discussed at https://oeis.org/A001599