Expand the following expression:
$$(2^0+2^1+2^2)(3^0+3^1)(5^0+5^1)(7^0+7^1)$$
Each combination of four choices (e.g. $2^2, 3^1, 5^0, 7^1$, whose product is $2^23^15^07^1=84$) gives a distinct factor of $420$.
PARTIAL ANSWER
That every integer ratio $n$ can be achieved , is probably impossible to be proven. But in the range $[2,16]$ , there are solutions in form of $p^4$ , where $p$ is a prime number :
2 16 2
5 625 3
11 14641 4
13 28561 5
43 3418801 6
83 47458321 7
307 8882874001 8
463 45954068161 9
1597 6504586067281 10
4217 316238254381921 11
20747 185276879591884081 12
102829 111805314979382104081 13
328901 11702018374499428575601 14
1799797 10492865217042730623781681 15
6498419 1783326405035972793339192721 16
The smallest prime power solutions are
gp > for(k=1,10,n=1;gef=0;while(gef==0,n=n+1;if(ispower(n)>0,if(ispseudoprime(n)==0,if(omega(n)==1,if(omega(n-1)==k,gef=1;print(k," ",n)))))))
1 4
2 16
3 121
4 841
5 17161
6 175561
7 2042041
8 214007641
The smallest solutions with more than one prime factor (conjecture : all of them have $2$ prime factors) :
1 15 2
2 391 2
3 30031 2
4 9699691 2
Solutions for larger $\alpha$ :
- $\alpha(n)=17$ : $1109^{12}$
- $\alpha(n)=18$ : $1627^{12}$
- $\alpha(n)=19$ : $137^{24}$
- $\alpha(n)=20$ : $211^{24}$
UPDATE : Dirichlet's theorem guarantees that for infinite many positive integers $\ n\ $ , a prime number $\ p\ $ with $\ \alpha(p^2)=n\ $ exists. And if the Schinzel hypothesis is true , such a prime exists for every positive integer $\ n\ $ , hence the above conjecture is true.
Proof :
Define $$t:=\prod_{j=2}^n p_j$$ where $\ n\ge 2\ $ and $\ p_j\ $ is the $j$-th prime number. Dirichlet's theorem guarantees that there is a positive integer $\ s\ $ such that $\ q:=st+1\ $ is prime. Then $$\omega(q^2-1)\ge \omega(q-1)=\omega(st)\ge \omega(t)=n-1$$ Hence $\ \omega(q^2-1)\ $ with prime $\ q\ $ is unbounded. Therefore infinite many $\ n\ $ must be realizable.
Assume that with the above $t$ , we can find a prime $\ p>p_n\ $ such that $\ 2tp+1\ $ and $\ 4tp+1\ $ are both prime. This is possible if the Schinzel hypothesis is true. Then with $\ q:=4tp+1\ $ we have $$\omega(q+1)=\omega(4tp+2)=\omega(2(2tp+1))=2$$ and $$\omega(q-1)=\omega(4tp)=n+1$$ Because of $\ \gcd(q-1,q+1)=2\ $ , we get $$\omega(q^2-1)=\omega(q-1)+\omega(q+1)-1=n+2$$
hence every positive integer $\ n'=n+2\ge 4\ $ is realizable.
Best Answer
If $p$ is an odd prime , the number $3^p-1$ cannot be a power of $2$ (Catalan's conjecture , now proven , but it might be easier to prove this particular claim), hence $3^p-1$ has an odd prime factor.
Moreover, if $p$ and $q$ are distinct odd primes, we have $\gcd(3^p-1,3^q-1)=3^{\gcd(p,q)}-1=2$ , hence if we choose an odd prime factor of $3^p-1$ and an odd prime factor of $3^q-1$, they must be distinct.
Hence for every odd prime factor of $n$, we have an odd prime factor of $3^n-1$ without duplicates. Since $2$ is always a factor of $3^n-1$, we have shown $\omega(3^n-1)\ge \omega(n)$.
To complete the proof, we have to show that $3^{2p}-1$ has two distinct odd prime factors , if $p$ is an odd prime , but this follows from $\gcd(3^p-1,3^p+1)=2$ and the fact that $3^p+1$ can also be no power of $2$.