$5$-Smooth Pairs

$5$-smooth numbers are numbers whose largest prime factor doesn’t exceed $5$.
$5$-smooth numbers are also called Hamming numbers.
Let $\Omega(a)$ be the count of prime factors of $a$ (counted with multiplicity).
Let $s(a)$ be the sum of the prime factors of $a$ (with multiplicity).
For example, $\Omega(300) = 5$ and $s(300) = 2+2+3+5+5 = 17$.
Let $f(n)$ be the number of pairs, $(p,q)$, of Hamming numbers such that $\Omega(p)=\Omega(q)$ and $s(p)+s(q)=n$.
You are given $f(10)=4$ (the pairs are $(4,9),(5,5),(6,6),(9,4)$) and $f(10^2)=3629$.
Find $f(10^7) \bmod 1\,000\,000\,007$.

Hamming numbers’ are 5-smooth numbers and can be written as 2^p*3^q*5^r, where p, q and r are nonnegative integers. You are given a function f(n) which calculates pairs of Hamming numbers that satisfy certain conditions. You need to find f(10^7) modulo 1,000,000,007.

I would suggest you to consider the prime factorizations of p and q. You need to find pairs (p, q) such that Ω(p)=Ω(q) and s(p)+s(q)=n. This means that you need to choose exponents for 2, 3, and 5 in the prime factorizations of p and q that satisfy the above conditions. Additionally, since you need to calculate f(10^7) modulo 1,000,000,007, you likely need to utilize properties of modulo arithmetic to avoid overflow.

It’s also notable that the problem provides you with f(10)=4 and f(10^2)=3629. These pieces of information might be useful for you to observe some patterns.

Mapping out all possible pairs and relating those to ω(x) and s(x) could also prove to be helpful in solving this problem. Exploring the relationships between these numbers and trying to establish a pattern could be the way forward.

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