Let $f(n)$ be the number of groups of order $n$ up to isomorphism. We want to prove that: $$f(a) \cdot f(b) \leq f(a \cdot b)$$ for all nonnegative integers $a$ and $b$. Our progress:
If $a \cdot b \leq 8191$, then our conjecture holds.
If $a + b \leq 848$, then our conjecture holds.
If $\gcd(a, b) = 1$, then our conjecture holds.
If $a \cdot b$ has less than 5 primes in its integer factorization, then our conjecture holds.
If $f(a \cdot b) \leq 3$, then our conjecture holds.
Let $c$ and $d$ nonnegative integers. Let $p$ a prime number. If $c + d \leq 9$ or $c=d$ or $c\cdot(5\cdot c + 24) + d\cdot(5\cdot d + 24) \leq 17 \cdot c \cdot d + 9$ then $$f(p^{c}) \cdot f(p^{d}) \leq f(p^{c + d})$$
Methods:
A lower bound of $f(2048)$ was used. What are the tightest known bounds for the number of groups of order $2048$?
For all positive integers $a$ and $b$: $$f(a) \leq f(a \cdot b)$$
If $a \cdot b = 0$, then our conjecture holds. If $1\leq a \cdot b \leq 8191$: 37592 tests in an algorithm.
180625 tests in an algorithm. If $f(65274) = 36$, then $36 \leq f(130548)$.
Thanks @testaccount.
Let $p$, $q$, $r$, $s$ distinct prime numbers. Study of 11 cases for $a \cdot b$: $p$, $p^{2}$, $p\cdot q$, $p^{3}$, $p^{2}\cdot q$, $p \cdot q \cdot r$, $p^{4}$, $p^{3} \cdot q$, $p^{2} \cdot q^{2}$, $p^{2} \cdot q \cdot r$, $p \cdot q \cdot r \cdot s$.
Our conjecture is valid for all squarefree $a \cdot b$.
Let $p$ a prime number and $m$ a positive integer.
$$p^{\frac{2}{27}m^{2}(m-6)} \leq f(p^{m}) \leq p^{\frac{1}{6}(m^{3}-m)}$$
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