Abundant numbers of form n^a+n+1

[Martin Fuller]

--- Dean Hickerson <dean at math.ucdavis.edu> wrote:

> 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
> 

Another solution is n=19581212842 which has
n^2+n+1 = 3*7^2*13*19*31*37*43*61*79*97*103*4447
sigma(n^2+n+1)/(n^2+n+1) = 2.0031147...

This is the smallest solution with n^2+n+1 divisible by any 6 of
{3,7,13,19,31,37,43}.

Martin Fuller