Propose $J(x)$, which is a function that takes in a number and outputs the sum of all its factors (including itself)
Firstly, I think it's pretty interesting as it allows you to describe certain types of numbers as solutions to different equations using this function. For example:
Perfect numbers are the solution to $J(x) = 2x$
Prime numbers are the solution to $J(x) = x+1$
But now to the problem that I need to ask you guys - as I sieved through data I saw that a lot of different inputs gave the same outputs, such as $J(6) = J(11)$ (they both equal 12)
I wondered what the difference between values that give the same output are, or in more mathematical language: $J(x+a) = J(x)$
I conjectured that for any integer $a$, there is at least one integer solution for $x$.
Some examples:
When $a$ is equal to $1$, a possible solution for $x$ is $14$
When $a$ is equal to $2$, a possible solution for $x$ is $33$
When $a$ is equal to $3$, a possible solution for $x$ is $382$
When $a$ is equal to $69$, a possible solution for $x$ is $8786$
So I guess my question to you guys is can you prove/disprove it?
Link to rudimentary python program for finding a specific gap value: https://repl.it/@JakubSkop/Factor-addition-function-gapvalues
Well, here's a start. Let $S$ be the set of positive integers $a$ such that $\sigma(x) = \sigma(x+a)$ has a solution. Since $\sigma$ is multiplicative, if $\sigma(x) = \sigma(x+a)$ and $r$ is coprime to both $x$ and $x+a$, then $\sigma(rx) = \sigma(r(x+a))$, so $ra \in S$.
Thus since $\sigma(14)=\sigma(15)$ for $a=1$, all $r$ coprime to $2, 3, 5$ and $7$ are in $S$. That's a set of positive asymptotic density ($(1-1/2)\cdot(1-1/3)\cdot (1-1/5) \cdot (1-1/7) = 8/35$).
Since we know that $2,3, 5$ and $7$ are in $S$, all primes are in $S$.
Similarly, since $\sigma(54)=\sigma(56)$ for $a=2$, if $r$ is coprime to $2, 3, 7$ then $2r \in S$.
And since $\sigma(16) = \sigma(25)$ for $a=9$, if $r$ is coprime to $2$ and $5$ then $9r \in S$.