TITLE: Classifying all numbers $n$ with the property that if $p$ is a prime, then $p \mid n \iff p-1 \mid n$
QUESTION [19 upvotes]: If a number $n$ has the property that if $p$ is a prime, then $p \mid n \iff p-1 \mid n$, we call $n$ a nice number for brevity.
A recent question on MSE (edit: now deleted) asks to prove that $1806$ is the only squarefree nice number. (That question gives no context and looks like a contest problem, and was therefore ill-received, but the actual question is a nice puzzle.) After solving it, I relayed the question to some friends, but initially I accidentally forgot the squarefree condition. This led me to wondering about nice numbers that are not necessarily squarefree.
A fairly simple argument proves that all nice numbers must be multiples of $1806$, and this is the first step towards solving the puzzle as well. (Edit: namely, for any nice $n$, we have $2-1 \mid n$, so $2 \mid n$; but then also $3 \mid n$. It follows that $6 \mid n$, and therefore $7 \mid n$. Then $2 \cdot 3 \cdot 6 = 42 \mid n$, so $43 \mid n$, and we obtain that $2 \cdot 3 \cdot 6 \cdot 43 = 1806 \mid n$ -- but $1807$ is not prime, so the argument ends here.)
However, there are more examples that are not squarefree:

$1806 = 2 \times 3 \times 7 \times 43$.
$12642 = 2 \times 3 \times 7^2 \times 43$.
$88494 = 2 \times 3 \times 7^3 \times 43$.
$6030842622 = 2 \times 3 \times 7 \times 43^2 \times 77659$.

These are all examples up to $2 \times 10^{10}$ (via computer search). The sequence is not currently listed on OEIS. The sequence has now been listed, at A345765.
My question is: can we classify all nice numbers? Are there only four? Are there more? Infinitely many?

REPLY [9 votes]: Summary

This is not a complete answer, but I think it's about as far as is both
feasible and interesting.
I've started calling these numbers saturated rather than nice, and will
do so for the rest of this answer.
There are a lot more saturated numbers than listed in the question and
comments, but they are nonetheless scarce: there are 126 below $10^{1000}$.
I now conjecture that there are infinitely many.

Downward/upward saturated
If $n$ has the property that for any prime $p$,
$$
    p \mid n \implies p-1 \mid n,
$$
we call it downward saturated. (This is equivalent to: $\lambda(n) \mid n$, where $\lambda$ is the Carmichael function. This sequence is A124240 on OEIS.) If it has the property that for any prime $p$,
$$
    p - 1 \mid n  \implies p \mid n
$$
we call it upward saturated. Clearly $n$ is saturated if and only if it is
both downward and upward saturated.
Suppose $n$ is already downward saturated. Then we can define
$$
n' = n \times \prod \{p\text{ prime} \mid p \not \mid n, (p-1) \mid n\}.
$$
Note that:

$n \mid n'$;
$n'$ is still downward saturated: the primes $p$ that we added to $n'$ by
construction have $p-1 \mid n$, and thus $p-1 \mid n'$;
if $n \mid m$ and $m$ is saturated, then $n' \mid m$;
in particular, $n = n'$ if and only if $n$ is saturated.

This gives a recipe for generating saturated numbers up to a cutoff $C$: start
with any downward saturated number $n$, and define the sequence $n, n', n'',
\ldots$. If it eventually stabilizes, you have found a saturated number. If the
sequence eventually exceeds $C$, then $n$ is not the divisor of any saturated
number below $C$.
Note that the sequence $n, n', n'', \ldots$ typically grows very fast when it
does not immediately stabilize: this allows us to quickly rule out $n$ and
search up to quite large cutoffs $C$.
It remains to be seen how we can find the right 'seeds' $n$ to start running
this generation algorithm.
Bases and saturation primes
For a fixed cutoff $C$, the (C-)saturation primes are those primes which
occur as a prime factor of a saturated number below $C$. Note that we are
currently unable to rule out any prime factor from being a saturation prime for
some $C$.
If we know what the saturation primes are for a cutoff $C$, say $p_1, \ldots,
p_k$, we are essentially done: we simply generate all numbers of the form
$p_1^{e_1} \cdots p_k^{e_k}$ that are below $C$, check whether they are
saturated, and we are done. Thus we need a procedure which finds these primes
for us.
Suppose we have found a partial list of the saturation primes $p_1, \ldots,
p_l$ (may be empty, but we may as well initialize with $2, 3, 7, 43$), and
there is a saturated number $K$ which is divided by a different prime. Let $q$
be the least prime divisor of $K$ which is not yet among our list. Then the
prime divisors of $q-1$ are among $p_1, \ldots, p_l$.
This means we have two strikes with one stone: we will enumerate all products
of powers of $p_1, \ldots, p_l$ that are below the cutoff. I call such numbers
bases. For each basis $n$, we construct the sequence $n, n', n'', \ldots$ as
above, until it exceeds the cutoff or until it saturates. Now note that $q - 1$
is a basis itself: this means it must be part of our enumeration, and through
it some saturated number $s$ such that $q-1 \mid s \mid K$. We can then add $q$
to our list of primes, and continue our search.
What remains are just tricks to speed up the computation:

Short-circuit past bases that are not downward saturated.
If the saturation sequence of a basis exceeds the cutoff, then so does the saturation sequence of each multiple of that basis, so we short-circuit past these as well.
Rather than using the sequence $n, n', n'', \ldots$ as above, we only add one
prime at a time, to avoid creating gargatuan numbers with an infeasible
number of divisors.
Keep track of each large number's factorization, so that we never have to
factorize.

Results
I have put my code implementing the above on Github: https://github.com/mjdv/saturated-numbers
With that code, I have listed all saturated numbers up to $10^{1000}$: https://pastebin.com/MkEfkSpY
I have also listed all saturated numbers up to $10^{1500}$, under the assumption that $7^4$ does not divide a saturated number below $10^{1500}$ (see below): https://pastebin.com/CT7hUDXi
There are 16 saturation primes for cutoff $10^{1000}$. They are
$$
2, 3, 7, 43, 77659, 21108889701347407, 5474088843701260097485589623,
5474159333397668466502066699,
14409061174110271629491692889111901658580261328754207,
12187898054314878179186265415000535659762253573563119365846458063397719,
4432982211548497951181997741316103259463454892983224005155108393681585716520...\\ 47036741930999,
5608017333924641389333679842752042439016754139199524596804923772035677581183...\\ 43740900139639,
1872705977981591020313336244505607714217889071693212286852076373413830346691...\\ 635632706249598340007,
9724040345982427116322290520909526374336288011833033457100867657726910715963...\\ 7823909035022095182893334450446173318074919,
3261106282862834686088755611243542924322840903892344944538844883173926466304...\\ 3468198518796043304218142825248440850700658152098958744500089400281673745399...\\ 0562312935495319926909431192160160253334732061914415799939878042598694784290...\\ 509218689159009145647650418769156819250302503,
3758874354521422972974642113256353902974703838722946991775778945748849417310...\\ 8680865452438274881828114249910402060654847053902299762834024507515311159860...\\ 0956871660411763936550716357782852055668471710382751800336614044465508091526...\\ 9963990635230724828608976209217399558456147493114652686096883786754935980432...\\ 41430009359.
$$
Edit: here are the same primes, in a compact notation suggested by @Vepir below: the primes are numbered $p_1, \ldots, p_n$, and a vector $(e_1, \ldots, e_k)$ stands for $p_1^{e_1} \cdots p_k^{e_k} + 1$.
$$
(),
(1,), 
(1, 1), 
(1, 1, 1), 
(1, 1, 1, 2), 
(1, 1, 1, 9), 
(1, 1, 1, 3, 1, 1), 
(1, 1, 1, 10, 2), 
(1, 1, 1, 1, 10), 
(1, 1, 1, 1, 3, 0, 0, 0, 1), 
(1, 1, 1, 1, 7, 0, 0, 0, 1), 
(1, 1, 1, 7, 1, 1, 2), 
(1, 1, 1, 8, 2, 1, 2), 
(1, 1, 1, 1, 13, 0, 0, 0, 1), 
(1, 1, 1, 9, 1, 1, 2, 0, 0, 0, 0, 2), 
(1, 1, 1, 1, 6, 0, 0, 0, 1, 2, 1).
$$
It seems to work as follows: once a saturation sequence incorporates a small
prime (apart from the few special ones in this list), it will no longer
stabilize. However, if it incorporates a large prime, this is fairly frequently
fine: the large prime does not cause a snowball effect. In fact, the number of
'safe' large primes grows fairly steadily. Up to cutoff $10^{1500}$, there are
at least 4 more. This leads me to the conjecture that there are infinitely many
saturated numbers.
There are two obstructions to searching larger ranges for saturated numbers:

For large cutoffs, it is slow to check that the saturation sequence of $2
  \times 3 \times 7^4 \times 43$ diverges past the cutoff. I have checked this
up to $10^{1000}$, and it takes up the majority of the computation time (hours,
compared to minutes for the rest of the computation). The code linked above
hard-codes this case as impossible, which allows listing all the other numbers up to $10^{1500}$ without too much patience.
Primality checking of very large numbers is somewhat slow. This seems to be a
fairly impenetrable barrier.

Some notes:

It seems extremely unlikely that $4$ divides any saturated number (and therefore any prime of the form $4k + 1$). If it does, that saturated number must have more than 10,000 distinct prime factors.
With the all the computational short-cuts, even for very large bounds, there are only a few-hundred paths $n, n', n'', \ldots$ that we follow through to either saturation or until they exceed the cutoff. Most are ruled out by earlier computations.