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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.




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