hi again,
>From: Jud McCranie <[EMAIL PROTECTED]>
>To: Alan Simpson <[EMAIL PROTECTED]>
>CC: [EMAIL PROTECTED]
>Subject: Re: Mersenne: this 3/2 conjecture and a result of Wagstaff
>Date: Wed, 23 Jun 1999 10:54:19 -0400
>
>There's no heuristic argument that I know of for 3/2 (it just fits known
>data
>well). Wagstaff's equation is a vast overestimate for small n, but maybe
>better for large n, or an upper bound.
>
If Wagstaff's heuristic is valid (and number theorists do seem to find it
convincing --- take that statement with handful of caution, after all
mathematics is about proofs not heuristics), then asymptotically the n-th
Mersenne prime should get closer and closer to e^(gamma*n). Again, if true,
this means that the 3/2 conjecture would become a progressively worse and
worse approximation.
As a number theorist, I would guess that the reason why e^{gamma*n) is not
such a good approximation (is this really true?) for the Mersenne primes
that we have seen so far is because of lower order terms.
It is clearly not the case that the exponent of the n-th Mersenne prime is
not (3/2)^P{n} or e^(gamma*n), but something like c^{n+o(n)), where "o(n)"
is the usual "little-o of n" (lim_{n \rightarrow \infty} o(n)/n = zero (a
severe abuse of notation in that limit!).
Do we have enough data to make any sensible guesses about the nature of this
"o(n)" term?
And another question, how does this linear curve (the term in the
exponential is linear in n, I mean) that people seem to want to attach to
the growth of the exponent of the n-th Mersenne prime change as n grows.
Could it be that if you look at the first 5 primes, and then the first 10
primes, etc., that the slopes are changing in some consistent manner?
Perhaps going to some limit? I guess this last question gets back to Jud
McCraine's (and others') point that it looks like (3/2)^{n}.
Alan Simpson
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