Does $D(n)$ always depend on $\gcd(n,\sigma(n))$ when $\sigma(N)=aN+b$, $N=xn$, and $n$ is a square?

Asked May 10 '18 at 14:56

Active May 11 '18 at 04:51

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Does $D(n)$ always depend on $\gcd(n,\sigma(n))$ when $\sigma(N)=aN+b$, $N=xn$, and $n$ is a square (with $\gcd(x,n)=1$)?

Here, $\sigma$ is the sum of divisors and $D(n) := 2n - \sigma(n)$ is the deficiency of $n$.

Copied from the comments For odd $N$ satisfying $\sigma(N)=2N$, we have $N=xn$ where $n$ is a square and $\gcd(x,n)=1$. In this case, it is known that $D(n)$ is a function (more accurately, a multiple) of $\gcd(n,\sigma(n))$.

edited May 11 '18 at 04:51

asked May 10 '18 at 14:56

Jose Arnaldo Bebita Dris

8,471

For odd $N$ satisfying $\sigma(N)=2N$, we have $N=xn$ where $n$ is a square and $\gcd(x,n)=1$. In this case, it is known that $D(n)$ is a function (more accurately, a multiple) of $\gcd(n,\sigma(n))$. – Jose Arnaldo Bebita Dris May 10 '18 at 15:13
More precisely: If $N=q^k n^2$ is an odd perfect number with special prime $q$, then $$D(n^2) = \sigma(q^{k-1})\gcd(n^2,\sigma(n^2)).$$ – Jose Arnaldo Bebita Dris Feb 25 '19 at 21:40
For the case when $I(m) = \frac{\sigma(m)}{m} = \frac{p+2}{p}$ (where $p$ is an odd prime), it is known that $D(m) = (p - 2)\cdot\gcd(m,\sigma(m))$. (See this MSE question for more information.) – Jose Arnaldo Bebita Dris Jun 30 '22 at 02:18

Does $D(n)$ always depend on $\gcd(n,\sigma(n))$ when $\sigma(N)=aN+b$, $N=xn$, and $n$ is a square?

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