Let $H_n=\frac{a_n}{b_n}$ where $H_n$ is a n:th harmonic number and $a_n$ and $b_n$ are coprimes.
1/ If $n$ is a prime power, show that $b_n > b_{n-1}$
2/ Find the integer factorization of $b_{21}$
My progress so far is:
1/ By induction $H_{n}=H_{n-1}+\dfrac{1}{n}=\dfrac{a_{n-1}}{b_{n-1}}+\dfrac{1}{n}=\dfrac{na_{n-1}+b_{n-1}}{nb_{n-1}}$
Let $n=p^k$ where $p$ is a prime and $k$ an integer which makes $n$ a prime power.
$H_{n}=\dfrac{p^ka_{n-1}+b_{n-1}}{p^kb_{n-1}}$
When you shorten this fraction, I want to show that the $\mathit{denominator}>b_{n-1}$. This is how far I have reached. How do I proceed?
2/This can easily be solved with a computer but the purpose is to do it using the harmonic number. I don't know how to do that in an easy way.
Thanks!
For $1)$ you don't need induction.
First observe that the common denominator for the sum in $H_n$ is $LCM \{1,2,3,..,n \}$. Now, after you bring the fraction to common denominator, it might cancel, but then, the denominator gets smaller.
From here you can deduce that for all $n$
$$b_n \mid LCM \{1,2,3,..,n \} \,.$$
Now, assume that $n=p^k$. Then
$$b_{n-1} \mid LCM \{1,2,3,..,n-1 \} \,.$$
As the power of $p$ in $LCM \{1,2,3,..,n-1 \}$ is exactly $k-1$, it follows that the power of $p$ in $b_{n-1}$ is at most $k-1$.
Write $b_{n-1}=p^\alpha m$ with $\alpha < k$ and $p \nmid m$.
Then
$$\frac{a_n}{b_n}=\dfrac{p^k a_{n-1}+p^\alpha m}{p^k p^\alpha m}=\dfrac{p^{k-\alpha} a_{n-1}+ m}{p^k m} \,.$$
Or
$$p^k m a_n =b_n ( p^{k-\alpha} a_{n-1}+ m) \,.$$
Now $p^k$ is relatively prime with $ p^{k-\alpha} a_{n-1}+ m$ (Why?) thus it must divide $b_n$.
$m$ is also relatively prime with $ p^{k-\alpha} a_{n-1}+ m$ (Why?) thus it must divide $b_n$.
Hence $p^k$ and $m$ are two relatively prime numbers, both dividing $b_n$.
Therefore $p^km \mid b_n$.
This proves that
$$b_n >b_{n-1} \,.$$
Actually you can prove something stronger in this case:
As $$p^k m a_n =b_n ( p^{k-\alpha} a_{n-1}+ m) \,.$$
$$b_n \mid p^k m a_n \,.$$
But $b_n$ is relatively prime with $a_n$ therefore $b_n \mid p^km$.
This shows that in this case $$b_n =p^k m$$ which might be helpful for $2)$.
As a side result, whenever $n$ is a power of a prime, $n \mid b_n$.
Here are some ideas for 2)
First, the only prime factors of $b_{21}$ can be $2,3,5,7,11,13,17,19$. Also, $$b_{21} \mid \operatorname{lcm}\{1,2,3.., 21 \} = 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \,.$$
You know that $2^4 \mid b_{16}, 3^2 \mid b_9, p \mid b_p$ for all the other.
The following Lemma can be very helpful.
Lemma If $p^k \mid b_{n-1}$ and $p^k \nmid n$ then $p^k \mid b_n$.
Combining this Lemma with $p \mid b_p$ and $2^4 \mid b_{16}$ you get that $b_{21}$ must be divisible by $2^4 , 11, 13, 17$ and $19$.
Thus $$b_{21}=2^43^b 5^c 7 ^d \cdot 11 \cdot 13 \cdot 17 \cdot 19 \,,$$ with $$ b \in \{ 0, 1, 2 \} ; c,d \in \{0,1\} \,.$$
This reduces the problem to only finding the powers of $3,5,7$ in $b_{21}$.
Moreover you know that $3^2 \mid b_{17}$, $5 \mid b_{9}$ and $7 \mid b_{13}$. The question is if any of these powers decreases before $b_{21}$.
I am not sure, but I think you might be able to study recursively $a_{n} \pmod{9, 5, 7}$ up to $n=21$ to settle this issue.