Hello,


I would like to know if there is some kind of heuristics according to which the sequence $(S\_n)\_{n\in\mathbb{N}}$ would be a divisibility sequence, where $S\_n$ is the $n$-th superperfect number (i.e a number $m$ such that $\sigma(\sigma(m))=2m$).  

Thanks in advance.



Is the sequence of superperfect numbers a divisibility sequence?
The sequence of superperfect numbers is a dvisibility sequence if and only if no odd superperfect numbers exist.


To see this just note that the even superperfect numbers are known to be all powers of two, precisely they are $2^{k-1}$ such that $2^k -1$ is a (Mersenne) prime. 


Then, if all are even it is clearly a divisibility sequence, indeed each one divides *all* subsequent ones (not just those whose index is a multiple, as is the definition of a divisibilty sequence). 


By contrast, if an odd should exists its index would need to be a prime $p$, as otherwise it would be divisible by some preceding superpefect number--an even number; yet then the $2p$-th superperfect number can be neither odd, being divisible by the second superperfect number, nor a power of two, being divisible by the odd $p$-th superperfect number. 


It seems to be believed that there are no odd superperfect numbers (and thus the sequence is a divisibility sequence), and partial results in this direction (somewhat resembling those on odd perfect numbers). Various literature on this is available (freely); for example "On finiteness of odd superperfect numbers" by Tomohiro Yamada (see <http://arxiv.org/abs/0803.0437> ), showing in particular that for any fixed number of distinct prime divisors there can only be finitely many.


In brief, the heuristics for this being a divisibility sequence are literally the same as those for there not being an odd superperfect number, and there can be no additional/other heuristics. 


