[seqfan] more

[Tomasz Ordowski]

Hello!

Note that m^2 | (m+1)^m - 1 for every m > 0.
I noticed that if m = 2^n-1, then 2^m-1 | (m+1)^m - 1.
So if m = 2^n-1 and (m, n) = 1, then (2^m-1)m^2 | (m+1)^m - 1.
Hence, if m = 2^n-1 is squarefree, then (2^m-1)m^2 | (m+1)^m - 1.
In particular, this divisibility holds for Mersenne numbers. Let's define:
   Numbers M = 2^p-1 such that ((M+1)^M - 1) / ((2^M-1)M^2) is prime.
Such M must be a Mersenne prime. I found 2^3-1 = 7 and 2^7-1 = 127.
Maybe someone will find more (the "next" prime M = 2^127-1 is nice /;-).


Regards!

T. Ordowski
_____________
A060073 - OEIS <https://oeis.org/A060073> (a(n+1) = ((n+1)^n - 1)/n^2).
A127837 - OEIS <https://oeis.org/A127837> (cannot be Mersenne primes > 3).