Minimal Number of Distinct Prime Factors for Integer to have Abundancy Index Exceed Given Number

== Theorem ==
Let $r \in \R$.
Let $\mathbb P^-$ be the set of prime numbers with possibly finitely many numbers removed.
Define:
:$M = \min \set {m \in \N: \ds \prod_{i \mathop = 1}^m \frac {p_i} {p_i - 1} > r}$
where $p_i$ is the $i$th element of $\mathbb P^-$, ordered by size.
Then $M$ satisfies:
:$(1): \quad$ Every number formed with fewer than $M$ distinct prime factors in $\mathbb P^-$ has abundancy index less than $r$
:$(2): \quad$ There exists some number formed with $M$ distinct prime factors in $\mathbb P^-$ with abundancy index at least $r$
So $M$ is the minimal number of distinct prime factors in $\mathbb P^-$ a number must have for it to have abundancy index at least $r$.
For $r$ an integer greater than $1$:
If $\mathbb P^-$ is taken to be the set of all prime numbers, the values of $M$ are:
:$2, 3, 4, 6, 9, 14, 22, 35, 55, 89, 142, \cdots$
This theorem shows that this sequence is a subsequence of the sequence A256969 in the OEIS, only differing by an offset.
If we require the numbers to be odd, we remove $2$ from $\mathbb P^-$.
The sequence of values of $M$ are:
:$3, 8, 21, 54, 141, 372, 995, 2697, 7397, 20502, \cdots$
== Proof ==
First we show that abundancy index is multiplicative.
Let $n \in \N$ and let $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ be its prime factorization.
Then the abundancy index of $n$ is:
so abundancy index is multiplicative.
Next we show that $M$ exists.
Note that Sum of Reciprocals of Primes is Divergent.
By Divergent Sequence with Finite Number of Terms Deleted is Divergent:
:the sum of reciprocals of all elements of $\mathbb P^-$ is also divergent.
Observe that:
By Limit Comparison Test:
:$\ds \sum_{n \mathop = 1}^\infty \ln \frac {p_n} {p_n - 1}$ is divergent as well.
By Logarithm of Divergent Product of Real Numbers:
:$\ds \prod_{n \mathop = 1}^\infty \frac {p_n} {p_n - 1}$ diverges to infinity.
Hence:
:$\exists N \in \N: \forall n \ge N: \ds \prod_{n \mathop = 1}^\infty \frac {p_n} {p_n - 1} > r$
Therefore $\ds \set {m \in \N: \ds \prod_{i \mathop = 1}^m \frac {p_i} {p_i - 1} > r} \ne \O$.
By the Well-Ordering Principle, $M$ exists.
Finally, we prove our claims $(1)$ and $(2)$.
Let $n$ be a number formed with fewer than $M$ distinct prime factors in $\mathbb P^-$.
Let $n = q_1^{a_1} q_2^{a_2} \cdots q_k^{a_k}$ be its prime factorization, where $q_i \in \mathbb P^-$ and $k < M$.
Then:
This proves $(1)$.
Now define $M \bar \sharp = \ds \prod_{i \mathop = 1}^M p_i$ (an analog of the primorial for $\mathbb P^-$).
Consider the sequence of abundancy indices of $\paren {M \bar \sharp}^n$, where $n$ is a strictly positive integer.
We have:
This product is strictly increasing and tends to $\ds \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}$ as $n \to \infty$, which is strictly greater than $r$.
From the definition of convergence to a limit:
:$\ds \exists N \in \N: \forall n \ge N: \size {\frac {\map {\sigma_1} {\paren {M \bar \sharp}^n} } {\paren {M \bar \sharp}^n} - \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}} < \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1} - r$
Since $\ds \frac {\map {\sigma_1} {\paren {M \bar \sharp}^n} } {\paren {M \bar \sharp}^n} < \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}$ for all $n$:
:$\ds r < \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1} - \size {\frac {\map {\sigma_1} {\paren {M \bar \sharp}^N} } {\paren {M \bar \sharp}^N} - \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}} = \frac {\map {\sigma_1} {\paren {M \bar \sharp}^N} } {\paren {M \bar \sharp}^N}$
Therefore $\paren {M \bar \sharp}^N$ is a number formed with $M$ distinct prime factors in $\mathbb P^-$ with abundancy index at least $r$.
This proves $(2)$.
== Also see ==
* Even Integer with Abundancy Index greater than 9
Category:Abundancy