1
$\begingroup$

$\sigma$ and $\tau$ are the sum of divisors and divisor counting functions respectively. In general it does not seem to be the case that for any $n$, $\sigma(\tau(n))=\tau(\sigma(n))$ so it seems peculiar that this is true for the consecutive numbers 170, 171, and 172.

I was aware previously that 170 is the smallest number such that $\varphi(170)$ and $\sigma(170)$ are perfect squares.

Since the $\sigma$ function appears in both, I was wondering if one of these facts can be derived from the other.

Specifically, does 170 being the smallest number that makes $\varphi(170)$ and $\sigma(170)$ perfect squares enough to imply that 170, 171, and 172 must result in $\sigma(\tau(n))=\tau(\sigma(n))$? (or the other way around)

$\endgroup$
  • $\begingroup$ "Specifically, does 170... other way around)" well... sortof, yes, because $170$ is the only number with that property (being the smallest with $\phi$ and $\sigma$ perfect squares), and $170$, $171$ and $172$ have the desired property $\sigma(\tau(n))=\tau(\sigma(n))$, so logically, this would be true. Perhaps rephrasing that part is appropriate? $\endgroup$ – vrugtehagel Oct 12 '17 at 11:44
  • $\begingroup$ Haha, sorry you don't understand I can't make it clearer than that. $\endgroup$ – user207119 Oct 13 '17 at 3:54
0
$\begingroup$

Appears squarefree solutions to your problem are pretty rigid and not common, each prime factor $2,3$ or $p = 2^m 3^n - 1$

3 =  3 div 2 sigma  4  =   2^2
66 =  2 3 11 div 8 sigma  144  =   2^4 3^2
70 =  2 5 7 div 8 sigma  144  =   2^4 3^2
170 =  2 5 17 div 8 sigma  324  =   2^2 3^4
11811 =  3 31 127 div 8 sigma  16384  =   2^14
73830 =  2 3 5 23 107 div 32 sigma  186624  =   2^8 3^6
74730 =  2 3 5 47 53 div 32 sigma  186624  =   2^8 3^6
76398 =  2 3 7 17 107 div 32 sigma  186624  =   2^8 3^6
79662 =  2 3 11 17 71 div 32 sigma  186624  =   2^8 3^6
80454 =  2 3 11 23 53 div 32 sigma  186624  =   2^8 3^6
82390 =  2 5 7 11 107 div 32 sigma  186624  =   2^8 3^6
84490 =  2 5 7 17 71 div 32 sigma  186624  =   2^8 3^6
85330 =  2 5 7 23 53 div 32 sigma  186624  =   2^8 3^6
87890 =  2 5 11 17 47 div 32 sigma  186624  =   2^8 3^6
94605 =  3 5 7 17 53 div 32 sigma  186624  =   2^8 3^6
170130 =  2 3 5 53 107 div 32 sigma  419904  =   2^6 3^8
200090 =  2 5 11 17 107 div 32 sigma  419904  =   2^6 3^8
207230 =  2 5 17 23 53 div 32 sigma  419904  =   2^6 3^8
...
$\endgroup$
  • $\begingroup$ Interesting observation, thanks. $\endgroup$ – user207119 Oct 13 '17 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy