Abundant numbers of form n^a+n+1

[Dean Hickerson]

Hans Havermann wrote:

> Just a quick follow-up... I've got 20 solutions for abundant n^3+n+1,  
> from n=12210244 to n=271870216. In contrast, there are no solutions  
> for abundant n^2+n+1 up to n=10^9.

I missed the original message about this, so maybe you know this already.
It's easy to construct values of n for which n^2+n+1 is abundant:  List
the possible prime divisors of numbers of that form, namely those
congruent to 0 or 1 (mod 3); this is A007645:

3, 7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, 139

Now pick enough of these primes p so that the product of  1+1/p  is larger
than 2; e.g. we can take 3, 7, ..., 97.  For each of these, find an n
such that p divides n^2+n+1.  Use the Chinese remainder theorem to find
an n such that n^2+n+1 is divisible by all of them.  Using 3, 7, ..., 97,
the smallest such n is 139136064116422, for which

    sigma(n^2 + n + 1) / (n^2 + n + 1) = 2.00478683...

I doubt that this is the smallest n for which n^2+n+1 is abundant.  We
could use the same method with a different set of primes, or we could
also make n^2+n+1 divisible by p^2 for some of the primes, etc.  Probably
some such variation will give a smaller solution.  Finding the smallest
one may be difficult.

Dean Hickerson
dean at math.ucdavis.edu