Are there a closed form of near solutions to the equation: $2\sigma(n)=3n$?

Question

I would like to check the solution of this equation:

$$2\sigma(n)=3n$$ where $\sigma(n)$ is the sum divisor function.

Note: I know only $n=2$ is a theortitical solution, are there a closed form of near solution of the titled equation ?

for example until now i find this :$n={29}^{16}+1$

EDIT:I edited the question because it is related to the precedent question

Answer 1

For a near solution, you can look at $21$. We have $\sigma(21)=1+3+7+21=32$, so $2\sigma(21)=64=3(21)+1$. You won't get any less absolute error than that. You can find near solutions with much smaller relative error. Let $n$ be a product of large primes $n=pqr\dots z$ Then $\frac {\sigma(n)}n = 1+\frac 1p + \frac 1q +\frac 1r +\dots \frac 1z +\frac 1{pq}+\frac 1{pr}+\dots$ Start from some large prime (maybe your friend $\frac 12(29^{16}+1)$) and keep multiplying by successive primes until the right side is very close to $\frac 32$ As the sum of the inverses of the primes diverges, you can always exceed $\frac 32$ by taking enough of them, no matter how high you start. It will take a lot of them. You can tune it even closer by deleting the one that takes you just over $\frac 32$ and looking for the next larger prime that fits underneath $\frac 32$

For example, starting with the "large prime" $17$ I find $$17*19*23*29*31*37*41*43*47*53*59*61*67*71*73=1356299720254712758928449\\ \sigma(1356299720254712758928449)=2034799419064923979776000\\ 2*2034799419064923979776000-3*1356299720254712758928449=699677365709682766653$$ with a relative error of about$\frac {7 \cdot 10^{20}}{2*2034799419064923979776000} \approx 0.000017$ You can improve that by replacing $73$ with $79$, then adding in much larger primes to get close to the ratio $\frac 32$. I don't trust the precision in Excel that far.

Answer 2

No, there are no other solutions.

Let $n\ge 3$ be an answer for this equation. It is obvious that $n$ must be even. So $1$, $n/2$, $n$ are divisors of $n$ (and $n/2>1$). Sum of these is $3n/2 + 1 > 3n/2$. So there is no answer other than $2$.

Answer 3

Denote $\sigma_{-1} (n) = \sigma (n)/n$. Also, we need to define what "close" means. We look for such $n$ that $$f (n) := |\sigma_{-1} (n) - 3/2| < \varepsilon_n,$$ where $\varepsilon_n \to 0$ is predefined (depending o how close you want). We also note $n > 2$.

Let $n$ be a prime. Then, $\sigma_{-1} (n) = 1 + 1/n$ so, $f (n) = 1/2 - 1/n$.

Assume now $n$ is a prime power, i.e., $n = p^{\alpha}$ for a positive integer $\alpha$ and a prime $p > 2$. Then, $$\sigma_{-1} (n) = 1 + \frac {1} {n^{1/\alpha}} + \frac {1} {n^{2/\alpha}} \cdots + \frac {1} {n}$$ and $$f (n) = \frac {1} {2} - \left (\frac {1} {n^{1/\alpha}} + \frac {1} {n^{2/\alpha}} \cdots + \frac {1} {n}\right ).$$ Then, $f (3^\alpha) \to 0$ as $\alpha \to \infty$. Hence, $n = 3^\alpha$ is a solution and you get closer and closer as $\alpha$ gets bigger. For example, all $3^\alpha$ where $\alpha > N$ are solutions to your problem.

Now, suppose $n$ is squarefree, i.e, $n = p_1 p_2 \cdots p_k$. Then, $$\sigma_{-1} (n) = 1 + \left (\frac {1} {p_1} + \frac {1} {p_2} + \cdots + \frac {1} {p_k} \right) + \cdots + \frac {1} {n}.$$ But this gets larger than we want actually, and we can prove that $\sigma_{-1} (n) \to 3$. To see that, notice $\sigma_{-1} (p_1 \cdots p_{k + 1}) > \sigma_{-1} (p_1 \cdots p_{k})$. Hence, $f (n) \to \frac {3} {2}$ as $k \to \infty$.

Now, let $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$ for $\alpha_1, \alpha_2, \cdots > 1$. Then $\sigma_{-1} (n) > \sigma_{-1} (p_1 p_2 \cdots p_k)$, so no solution here, either.

So, the only solution is $n = 3^\alpha$.

Note that here I assumed you want to keep and actually improve the speed of your approximation with the growth of $n$. Otherwise, there are infinitely many scattered arbitrary solutions to the problem, which makes the problem ill-defined. For example, $n = 3 \cdot 11 \cdot 31 \cdot 257$ is a fine approximation.