This question relates to Robin's Inequality. Is $\sigma{(n^2)}$ < (2 n) $\sigma{(n)}$ ? For what integer values of n is this satisfied?
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suggest you get one of the easier computer languages, I suspect Python qualifies, where you can write subroutines for factoring by trial division (fine if the original number is no larger then, say, $10^{12}$), calculating $\sigma(n)$ and so on, all for numbers of modest size. Then run quite simple experiments on numbers, see what appears to work. If you arrive at some reasonable patterns but do not know how to prove that they work, or disprove by larger example, that would make for a better question. – Will Jagy Sep 18 '14 at 18:53
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Are there any number theory arguments that can determine this without 'math' experimentation? I am terrible at computer programming. – user128932 Sep 18 '14 at 20:23
3 Answers
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We can prove the inequality (actually a stronger inequality) with the explicit formula for $\sigma(n)$ in terms of the prime factorisation of $n$. For
$$n = \prod_{k=1}^r p_k^{\alpha_k},$$
we have
$$\sigma(n) = \prod_{k=1}^r \frac{p_k^{\alpha_k+1}-1}{p_k-1},$$
and so
$$\begin{align} \frac{\sigma(n^2)}{n\sigma(n)} &= \prod_{k=1}^r \frac{p_k^{2\alpha_k+1}-1}{p_k^{\alpha_k}(p_k^{\alpha_k+1}-1)}\\ &= \prod_{k=1}^r \frac{(p_k^{2\alpha_k+1}-p_k^{\alpha_k}) + (p_k^{\alpha_k}-1)}{p_k^{\alpha_k}(p_k^{\alpha_k+1}-1)}\\ &= \prod_{k=1}^r \left(1 + \frac{p_k^{\alpha_k}-1}{p_k^{\alpha_k}(p_k^{\alpha_k+1}-1)}\right)\\ &\leqslant \prod_{k=1}^r \left(1 + \frac{p_k-1}{p_k(p_k^2-1)}\right) \tag{$\ast$}\\ &= \prod_{k=1}^r \left(1 + \frac{1}{p_k(p_k+1)}\right)\\ &\leqslant \prod_{k=1}^r \left(1 + \frac{1}{p_k^2}\right)\\ &< \prod_{p\in\mathbb{P}} \left(1+\frac{1}{p^2}\right)\\ &= \prod_{p\in\mathbb{P}} \frac{1 - \frac{1}{p^4}}{1-\frac{1}{p^2}}\\ &= \frac{\zeta(2)}{\zeta(4)}\\ &= \frac{15}{\pi^2}\\ &< 1.52, \end{align}$$
where in $(\ast)$ we used that $\frac{m^\alpha-1}{m^\alpha(m^{\alpha+1}-1)} \leqslant \frac{m-1}{m(m^2-1)}$ for $m \geqslant 2$ and $\alpha \geqslant 1$, which can be elementarily verified. If we don't replace $\frac{1}{p(p+1)}$ with $\frac{1}{p^2}$ two lines below $(\ast)$, we get a sharp bound, which however I don't know how to evaluate explicitly. The sharp bound is approximately $1.368432778$.
Daniel Fischer
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So ($\sigma(n^2)$)/ $(n^2)$ < (1.52) ($\sigma(n)$)/n ; Thanks for the info. – user128932 Sep 19 '14 at 03:00
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To complete Daniel Fischer's answer:
A closed form for the sharp bound is $$\prod_{p \in \mathbb{P}}{\left(\dfrac{p^2 + p + 1}{p^2 + p}\right)} = \dfrac{\zeta(2)}{\zeta(3)} \approx 1.368432778.$$
See this MSE question for more.
Jose Arnaldo Bebita Dris
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Sorry: not sure about this. With the inclusion of the factor 2, your quantity is not multiplicative. Evidently the large (and increasing) values of $\sigma(n^2) / (n \sigma(n)),$ which really is multiplicative, occur at the primorials, that is the product of the consecutive primes. Here are the first few ratios:
n s(n) s(n^2) s(n^2) / n s(n)
1 1 1 1 1 = 1
2 3 7 1.166666666666667 2 = 2
6 12 91 1.263888888888889 6 = 2 * 3
30 72 2821 1.306018518518518 30 = 2 * 3 * 5
210 576 160797 1.329340277777778 210 = 2 * 3 * 5 * 7
2310 6912 21386001 1.339411037457912 2310 = 2 * 3 * 5 * 7 * 11
Should not be too hard to prove stuff and decide whether, at the primorials, the ratio is bounded. Simple analytic number theory bounds once you stick to primorials.
It appears that we always have $\sigma(n^2) < 2 n \sigma(n).$ As both sides are "multiplicative," it suffices to prove this for $1$ and all primes and prime powers. For example, prime $n=p,$ we get $\sigma(p^2) = p^2 + p + 1,$ while $n \sigma(n) = p (p+1) = p^2 + p,$ slightly smaller. But then double it and it becomes larger.
For $n=p^2,$ $\sigma(n^2) = \sigma(p^4) = p^4 + p^3 + p^2 + p + 1.$ Meanwhile, $n \sigma(n) = p^2 \sigma(p^2) = p^2 (p^2 + p + 1) = p^4 + p^3 + p^2,$ again slightly smaller, double it and it becomes bigger.
n s(n) s(n^2) s(n^2)-2ns(n) n factored
1 1 1 -1 1 = 1
2 3 7 -5 2 = 2
3 4 13 -11 3 = 3
4 7 31 -25 4 = 2^2
5 6 31 -29 5 = 5
6 12 91 -53 6 = 2 * 3
7 8 57 -55 7 = 7
8 15 127 -113 8 = 2^3
9 13 121 -113 9 = 3^2
10 18 217 -143 10 = 2 * 5
11 12 133 -131 11 = 11
12 28 403 -269 12 = 2^2 * 3
13 14 183 -181 13 = 13
14 24 399 -273 14 = 2 * 7
15 24 403 -317 15 = 3 * 5
16 31 511 -481 16 = 2^4
17 18 307 -305 17 = 17
18 39 847 -557 18 = 2 * 3^2
19 20 381 -379 19 = 19
20 42 961 -719 20 = 2^2 * 5
21 32 741 -603 21 = 3 * 7
22 36 931 -653 22 = 2 * 11
23 24 553 -551 23 = 23
24 60 1651 -1229 24 = 2^3 * 3
25 31 781 -769 25 = 5^2
26 42 1281 -903 26 = 2 * 13
27 40 1093 -1067 27 = 3^3
28 56 1767 -1369 28 = 2^2 * 7
29 30 871 -869 29 = 29
30 72 2821 -1499 30 = 2 * 3 * 5
31 32 993 -991 31 = 31
32 63 2047 -1985 32 = 2^5
33 48 1729 -1439 33 = 3 * 11
34 54 2149 -1523 34 = 2 * 17
35 48 1767 -1593 35 = 5 * 7
jagy@phobeusjunior:~$
Will Jagy
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GREAT info. So $\sigma(n^2)$ < (2 n) $\sigma(n)$ therefore ($\sigma(n^2)$)/ $(n^2)$ < 2 ($\sigma(n)$)/n. – user128932 Sep 18 '14 at 21:10
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@user128932, sorry, I need to think about this, after I run some errands, library and drugstore. Your idea may be true always, but a full proof is fairly elaborate. – Will Jagy Sep 18 '14 at 21:14
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1Not too bad. Using the explicit formula for $\sigma(n)$, you get $$\begin{align} \frac{\sigma(n^2)}{n\sigma(n)} &= \prod \frac{p_k^{2\alpha_k+1}-1}{p_k^{\alpha_k}(p_k^{\alpha_k+1}-1)} = \prod \left(1+\frac{p_k^{\alpha_k}-1}{p_k^{\alpha_k}(p_k^{\alpha_k+1}-1)}\right)\ &\leqslant \prod \left(1+\frac{p_k-1}{p_k(p_k^2-1)}\right) < \prod \left(1+\frac{1}{p_k^2}\right)\ &< \frac{\zeta(2)}{\zeta(4)} = \frac{15}{\pi^2} < 2. \end{align}$$ – Daniel Fischer Sep 18 '14 at 21:19
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@Daniel, thanks. Consider posting an answer, I am a little weary of this one, and relieved that there is a straightforward approach (I was going to do a big loop through primorials, not even guaranteed to work the way i had in mind). – Will Jagy Sep 18 '14 at 23:12