"Nathan Russell" <[EMAIL PROTECTED]>
Nathan Russell observes:
> This is something I've been toying with the last few days.
>
> I can prove that all even positive integers except 2, 4 and 8 can be written
> as the sum of Mersenne primes.
>
> All above 21 are either 0, 1 or 2 mod 3, and are therefore the sum of either
> 1, 2 or 3 sevens with a sufficient number of 3's thrown on top.
> I am sure this can (and should) be stated far more formerly, but my real
> question is this: Is it possible to strengthen this conjecture, say by
> putting a ceiling on the number of times that any one prime need be
> repeated?
>
No. Let e1 < e2 < ... be the exponents of the Mersenne primes.
If this sequence is finite, say ending at ek, then only finitely
many numbers are representable if we bound the number of summands.
Next suppose there are infinitely many Mersenne primes.
We can bound the partial sum
M(e1) + M(e2) + ... + M(ek)
< 2^e1 + 2^e2 + ... + 2^ek
< 1 + 2 + 4 + ... + 2^(ek) = 2^(ek+1) - 1.
To represent M(e(k+1)) as a sum of these, some summand must be repeated
at least 2^(e(k+1) - ek - 1) times. Both e(k+1) and ek are primes.
It is an easy exercise to show that there can be arbitrarily
large gaps between adjacent primes and hence
between adjacent Mersenne exponents.
Another proof starts with a bound m on the number of times each
summand is used. Using M(e1) through M(ek) we can form at most
(m + 1)^k different sums. Show that
M(e(k+1)) > (m+1)^k
for large k, whatever the choice of m.
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