In the same style as a previous test, I did a little test today looking for all the numbers such as the odd divisors, ordered in increasing order excluding $1$, when they are accumulated one by one to $1$ the resulting sums are always prime numbers:
The first elements of the sequence are $\{3,5,6,10,11,12,17...\}$, it is not included at OEIS:
$n=3$, odd divisors not including $1$ are: $\{3\}$ and $1+(3+1)=5 \in \Bbb P$.
$n=5$, odd divisors not including $1$ are: $\{5\}$ and $1+(5+1)=7 \in \Bbb P$.
$n=6$, odd divisors not including $1$ are: $\{3\}$ and $1+(3+1)=5 \in \Bbb P$.
For instance $n=39$ has $3$ divisors:
$n=39$, odd divisors not including $1$ are: $\{3,13,39\}$ and $1+(3+1)=5 \in \Bbb P$, $5+(13+1)=19 \in \Bbb P$, $19+(39+1)=59 \in \Bbb P$.
I tried to look for the one with the longest list of divisors, but it seems that the longest possible length is $7$ elements, for $n=34131$, at least up to $n=1.6 \cdot 10^9$ and in all cases $n$ is an odd number (thanks to @Peter for the feedback in the comments!):
$n=34131$, odd divisors not including $1$ are: $\{3,31,93,367,1101,11377,34131\}$ and $1+(3+1)=5 \in \Bbb P$, $5+(31+1)=37 \in \Bbb P$, $37+(93+1)=131 \in \Bbb P$, $131+(367+1)=499\in \Bbb P$, $499+(1101+1)=1601 \in \Bbb P$, $1601+(11377+1)=12979\in \Bbb P$, $12979+(34131+1)=47111 \in \Bbb P$.
So far heuristically I can see that the list of $n$'s includes some $p$ primes and $2p$.
I would like to ask the following question:
Why does it seem impossible to find a $n \gt 34131$ such as all the odd divisors comply with the property and the length of the list of those divisors is longer? Is there a theoretical reason behind? Is there a counterexample? (maybe a very big number due to the conditions of the property?)
Thank you very much in advance!