I was playing around and came up with the following formula: Let $d(x)$ be the divisor counting function and $(x,y)$ be the greatest common divisor of $x$ and $y$. If $n\in\mathbb{N}$ I write $$q_n={n^2+1 \above 1.5pt \left(n^2+1,d\left(n^2+1\right)\right)}$$
The formula seems circuitous but it appears that $q_n=$A228564; the largest odd divisor of $n^2+1$. It appears we have the following identities:
- $\left(q_{n+1},q_n\right)=$A109009
- ${q_n-1 \above 1.2pt 4}=$A123596
- ${q_n-1 \above 1.2pt 4}=$A061066 if $n$ is prime.
- ${q_{n+2}-q_n \above 1.2 pt 4}=$A026741
- ${q_{n+1}+q_n-1 \above 1.2pt 4}=$A001318
- ${q_{n+2}+q_n-1 \above 1.2pt 4}=$A147685
- ${q_{n+7}+q_{n+6}+q_{n+5}+q_{n+4}-q_{n+3}-q_{n+2}-q_{n+1}-q_n-1 \above 1.2pt 4}=$ A09225 for $n>3$
My question is straightforward:
Is $q_n$ the largest odd divisor of $n^2+1$ and if so why?
If this is not true a disproof or counter example suffices. I have only checked the first 50 or so terms. Note if this result is true and the second identity is true then for $n^2+1$ prime we have $n^2+1=4k+1$ where $k\in$A123596. Subtracting $1$ from both sides gives us $n^2=4k=$A075408. So $n=4\sqrt k$ and $k$ must be a square. In fact I have that $k$ is actually in a much smaller and sparse subset of A123596 namely $k\in$A230312. Suprisingly in a round about fashion using this "formula" I got the following result:
If $q_n$ and $n$ are prime and $\sigma_1\left(n^2+1\right)$ is the sum off all divisors of $n^2+1$ then $$\sigma_1\left(n^2+1\right)=12*A123596+6$$
The first counterexample is $n=443$, for which $\frac{n^2+1}{q_n} = 10$ is not a power of $2$.