Testwiki:Reference desk/Archives/Mathematics/2015 July 13

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July 13

Why is 18 a solitary number?

I couldn't find any information on the subject in google, except for some archive reference desk, which offered some proof that I failed to understand. Basically what I am asking is, why aren't there other numbers n except from 18, for which sigma(n):n = 13:6, where sigma(n) is the sum of the divisors of n including n itself. — Preceding unsigned comment added by 130.204.34.208 (talk) 07:51, 13 July 2015 (UTC)

The proof here is based on decomposing n and looking at the size of $\sigma (n)/n$. If $n=\prod _i{p}_i^{a_i}$ then $\sigma (n)=\prod _i\sigma \left(p_i^{a_i}\right)=\prod _i\frac{p_i^{a_i+1}-1}{p_i-1}$. Also if we let $f(n)=\sigma (n)/n$ we have $f(n)=\prod _{i}f\left(p_i^{a_i}\right)$. All the factors are greater than 1. If you try $n=2^a{3}^{b}m$ with $a=2,b=1$ the product $f(2^a)f(3^b)$ will already be greater than 13/6, and $f(m)$ will just make it larger, so it can't be right. If you try $a=1,b=2$ then the product will exceed 13/6 unless $m=1$ and then you have 18. The rest of the proof follows, using also the facts that $\sigma (6)=12$, and that for a prime $p>3$ we have $\sigma (p^a)$ is an integer and $p^a$ is odd. -- Meni Rosenfeld (talk) 10:01, 13 July 2015 (UTC)

And, since I just can't get enough of Inside Out, I'll add that 18 is solitary because it lost the core memory that powers friendship island. -- Meni Rosenfeld (talk) 13:21, 14 July 2015 (UTC)

I suppose they are a step up from The Numskulls. That cartoon strip has survived for fifty years now. Dmcq (talk) 10:45, 17 July 2015 (UTC)