# Any results on $\gcd(N^2, D(N^2))$ where $N^2$ is deficient and $D(N^2)$ is the deficiency of $N^2$?

Any results on $\gcd(N^2, D(N^2))$ where $N^2$ is deficient and $$D(N^2)=2N^2 - \sigma(N^2)$$ is the deficiency of $N^2$?

I checked OEIS sequence A033879 and have so far been able to get hold of Klyve, et. al's paper titled "On the difference between an integer and the sum of its proper divisors". Google does not appear to be of much help either.

It's an odd number between 1 and $N^2$, of course. It can actually be quite large: with $f(n)=\gcd(n^2,2n^2-\sigma(n^2)),$ $$f(26334) = f(2 \cdot 3^2 \cdot 7 \cdot 11 \cdot 19) = 3^2 \cdot 7^2 \cdot 11^2 \cdot 19^2 = 19263321.$$
Since $f(n)$ is odd, a somewhat tighter upper limit is $m^2$ where $m=n/2^\nu$ is the largest odd divisor of $n$. The above example has $f(n)=m^2/9,$ which shows that it would be hard to improve on this 'trivial' bound. On the other side, $f(n)=1$ infinitely often (at prime powers, for example).
• Thank you for your answer. Are other numbers known for which $f(n)=1$, if $n > 1$ and $n$ is not a prime power? – Jose Arnaldo Bebita Dris Oct 14 '16 at 3:37
• @ArnieBebitaDrisuser11235813 Right, though not all semiprimes. f(77) = 7, for example. But you can find infinitely many semiprimes n with f(n) = 1 using Dirichlet's theorem, I believe. And you can take it further, finding terms with $\Omega(n)$ arbitrarily large. – Charles Oct 14 '16 at 4:44