More problems in 2016 part 5

For $d$ prime numbers:

If $d-1=p_1 ^a\times p_2^b\times p_3^c\times\cdots$ , where $p_n$ are prime numbers.

$\large\begin{aligned} \cfrac{\phi(\phi(d))}{\phi(d)} &=\phi\left (\cfrac{\phi(d)}{d} \right )\\ &=\phi\left (\cfrac{d\left ( 1-\cfrac{1}{d} \right )}{d} \right )\\ &=\phi\left ( 1-\cfrac{1}{d} \right )\\ &=\phi\left ( \cfrac{d-1}{d} \right )\\ &=\cfrac{\phi\left ( d-1 \right )}{\phi\left ( d \right )}\\ &=\cfrac{(d-1)\left ( \cfrac{p_1-1}{p_1} \right )\left ( \cfrac{p_2-1}{p_2} \right )\left ( \cfrac{p_3-1}{p_3} \right )\cdots}{d\times\cfrac{d-1}{d}}\\ &=\left ( \cfrac{p_1-1}{p_1} \right )\left ( \cfrac{p_2-1}{p_2} \right )\left ( \cfrac{p_3-1}{p_3} \right ) \end{aligned}$

For $d$ composite numbers:

Let's $\Gamma(d)$ denote $\cfrac{\phi(\phi(d))}{\phi(d)}$

If $d=p_1 ^a\times p_2^b\times p_3^c\times\cdots$ :

$\large\Gamma(d)=\Gamma(p_1)^a\times \Gamma(p_2)^b\times \Gamma(p_3)^c\times\cdots$

Solution:

The divisors of $29233$ are: $\begin{array}{cccc} 1&23&31&41\\ 23\times31\times41&23\times31&23\times41&31\times41 \end{array}$

$1$ :

$1$

$23$ :

$\cfrac{1}{2}\times\cfrac{10}{11}=\cfrac{5}{11}$

$31$ :

$\cfrac{1}{2}\times\cfrac{2}{3}\times\cfrac{4}{5}=\cfrac{4}{15}$

$41$ :

$\cfrac{1}{2}\times\cfrac{4}{5}=\cfrac{2}{5}$

$23\times31$ :

$\Gamma(23)\times\Gamma(31)=\cfrac{4}{33}$

$23\times41$ :

$\Gamma(23)\times\Gamma(41)=\cfrac{2}{11}$

$31\times41$ :

$\Gamma(31)\times\Gamma(41)=\cfrac{8}{75}$

$23\times31\times41$ :

$\Gamma(23)\times\Gamma(31)\times\Gamma(41)=\cfrac{8}{165}$

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$\large \begin{aligned} \displaystyle\sum_{d\mid 29233}\cfrac{\phi(\phi(d))}{\phi(d)} &=1+\cfrac{5}{11}+\cfrac{4}{15}+\cfrac{2}{5}+\cfrac{4}{33}+\cfrac{2}{11}+\cfrac{8}{75}+\cfrac{8}{165}\\ &=\cfrac{5\times(11\times3\times5+5\times3\times5+4\times11+2\times3\times11+4\times5+2\times3\times5+8)+8\times11}{11\times3\times5\times5}\\ &=\cfrac{5\times(165+75+44+66+20+30+8)+88}{825}\\ &=\cfrac{2128}{825} \end{aligned}$

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So the solution is: $2128+825=2953$