I don’t think there is anything in this idea. Note that

  n! * n / (n-1)
  = n^2 * (n-2)!
  = n * (n-1 + 1) * (n-2)!
  = n! + n * (n-2)!
  = n! + (n-1 + 1) * (n-2)!
  = n! + (n-1)! + (n-2)!

So the conjecture here is that the length of the minimal superpermutation is

  n! + (n-1)! + (n-2)! + fib(2n - 6).

The only thing to be said in favour of this idea is that it lies between the known lower and upper bounds, so we can’t yet prove it’s wrong. But there is no reason at all to expect Fibonacci numbers to appear in this context.

The most plausible conjecture we have right now is that the length of the minimal superpermutation is

  n! + (n-1)! + (n-2)! + (n-3)! + (n-4)

for n > 3. We can’t prove this either, but at least there is reasonable evidence for it.

For n = 7, their formula gives 5901 and yours gives 5907. How feasible would it be to verify which one's correct? Exhaust every 5901-long sequence? Or is there some smarter way?

No one has yet managed to solve n=6 completely, so I think solving n=7 is a long way off.

On the other hand, I think we will soon have an improved lower bound that is strong enough to refute this formula.

I wonder if some day, some one will anonymously post solution to the P vs NP problem, without fully understanding the scope and effect of the solution on the world.

Some times not knowing the history and hardness of a tasks helps in a novice attempting an impossible task, and arriving at the solution, completely ignorant of the history of the problem.

"He didn't know it was impossible, so he did it".

That's one of best motto I've ever heard!

This almost certainly won't happen. First, because "P vs NP" is a very well-studied problem and many complex technical constraints have been proved about what kinds of things can and cannot be part of "P" algorithm for an NP-complete problem. There's no way to be educate enough to be able express something that could possibly be a well-formed proof (and not made meaningless by its vaguess) without knowing enough to navigate these concerns and therefore appreciating the magnitude of the problem.

Second, because the proof won't be useful outside of arcane theory. If P != NP, that verifies something everyone already asumes, so it wouldn't change much except for abstract curiosity of a few experts able to comprehend the complexity of the meaning.

Out of curiosity, if someone proved P to not equal NP, what kind of scope and effect would it bring to the world outside academia? P == NP would obviously be a world-transforming proof

The actual result of "P =/= NP" wouldn't actually be very interesting or informative - after all, we're all pretty sure of that anyway.

What would be very interesting - and likely give pretty deep insights into the very nature of computation - would be how exactly they proved it.

Right. The important thing about the Clay Mathematics Millenium Prize is that it is a set of well-chosen SMART-ish goals. They want to incentivize work on really hard fields of mathematics but they have tried to do this by choosing results that are each not too intimidating. The Navier-Stokes prize, for example, allows you to assume all of the nicest features of Navier-Stokes problems in practice -- basically all of the fluid flows are well below supersonic and are occurring in highly homogeneous, nice fluids -- and ask the most basic mathematical question: does a smooth solution to these equations always exist? That question by itself is not so important, but it's specific and measurable. But it's a bit of a "reach" -- you would have to have some cutting insight about turbulence in order to answer this question one way or the other. That cutting insight is what they're trying to incentivize.

P vs NP is the same: if solutions to a problem are easy to check, is there always some better way to analyze the mechanics of the checker to make those solutions easy to find, so that we aren't stuck with brute-forcing it? Whether the answer goes one way or the other, the point is that solving the problem would have to provide some insight to the effect of "here is a periodic table of elements for all of the 'easy' algorithms -- and here are the properties of all the 'molecules' made by combining those building blocks." And only once someone advances our understanding with those cutting insights can we say "yes we can always reach into the verification mechanism to understand it well enough to build a better-than-brute-force algorithm" or "no, here is such a thing that algorithms cannot do, it's not just that I am not smart enough to find a way for them to do this -- they are fundamentally incapable of doing it faster."

It would probably give us some new insight on what computation actually is. Our inability to formally proof class separation suggests we're missing something.

as I understand it would have little effect since it's implicitly assumed to be that way, might be wrong though

RSA algorithm is based upon factorization, reverse problem for which is in NP. If a solution's difficulty is a low degree polinom (3 or lower, but I might be wrong) it could be used to get private keys from public ones in a matter of days.

The proof itself would not give factorization algorithm.

Even if P = NP, prime factorization can be very time consuming to solve. The prime factorization problem is in the NP class, but we don't know if it is NP-hard.

Factorization is not NP, it's "easier"

It's definitely in the class, but it's undecided which subclass it belongs to.

See https://en.wikipedia.org/wiki/Huffman_coding#History

relevant story: https://en.wikipedia.org/wiki/George_Dantzig#Mathematical_st...