The Set Up:
I was studying this sequence which can be described as follows.
Call the linked sequence $a(n)$. We define $b(n)$ a related sequence first linked here:
$$ b(n) = \text{largest k} | \exists m > 1 | \sigma_1(n) = m^k $$
Where $\sigma_1(n)$ is the sum of the factors of $n$.
We can then define:
$$ a(1) = 1 $$
$$ a(n+1) = \text{smallest r} | b(r) > b(a(n)) $$
The Observation/Question:
One can easily verify that every element of $a(n)$ listed in OEIS is a product of mersenne primes. The question naturally arises, is this always true? Why might this be the case?
Some Attempts to Prove:
To start off we can consider a number that is a product of 2 mersenne primes, call those primes $p_1, p_2$ which are equal to $2^{k_1} - 1, 2^{k_2} - 1$ respectively.
Then the sum of divisors of the product $p_1 p_2 $ would be $1 + (2^{k_1} - 1) + (2^{k_2} - 1) + (2^{k_1} - 1)(2^{k_2} - 1) = 1 + 2^{k_1} - 1 + 2^{k_2} - 1 + 2^{k_1 + k_2} - 2^{k_1} - 2^{k_2} + 1 = 2^{k_1 + k_2}$
And this clearly has a massive exponent. It suggests that products of mersenne primes, will have sum of factors of the form $2^l$, and will be the smallest numbers that have sum of factors are of the form of an exponent $l$, and therefore will be the FIRST numbers in $b(n)$ to break the threshold of $l$.
So essentially what is left to be shown is that:
- We can easily scale our inclusion and exclusion formula to arbitrarily many mersenne primes
Some Further Notes:
We can observe that given an expression $2^l$ we can partition $l$ into a partitions of the form $c_1,c_2.. c_r$ where $c_1 + c_2 + ... c_r = l$. Then the number $(2^{c_1} -1)(2^{c_2} -1) ... (2^{c_r} - 1)$ would have sum of divisors $2^l$.
There should be some way to select the $c_i$ which minimizes this, and essentially the values of $a(n)$ correspond to the minimizing partitions here (by way of prime factorization)