Abundant numbers of form n^a+n+1

[Jack Brennen]

Out of curiosity, I began a search for abundant numbers of
the form X = n^a+n+1.  (n>=2, a>=2).

The findings are somewhat interesting.  Here are the smallest
10 such numbers X, along with their ratios sigma(X)/X (which
must be over 2 for any abundant number):

238^4+238+1 == 3208542975 2.015323731171155655161514550
172^5+172+1 == 150536645805 2.022798549626552176386191535
562^5+562+1 == 56063676973395 2.107841704303467796858806830
592^5+592+1 == 72712409055825 2.256986488702412943961020428
802^5+802+1 == 331796531264835 2.015381957296763658265346953
877^5+877+1 == 518797610149035 2.095235478169409031223220410
4963^4+4963+1 == 606704338819125 2.009758093308633804129096791
5188^4+5188+1 == 724435742643525 2.123799116793552924070702263
1087^5+1087+1 == 1517566463015295 2.011913630367850176340859037
1192^5+1192+1 == 2406474571744425 2.041692387939998367058484374

The interesting part is that only a=4 and a=5 are represented
up to this search limit.  The polynomials n^2+n+1 and n^3+n+1
are much "denser" up to this limit, but yield no abundant
numbers yet.

It would be interesting to know how the density of abundant
numbers in n^a+n+1 varies for different values of a.  Clearly,
a=2 and a=3 are in a different class than a=4 and a=5.
One thing that is easy to see is that neither x^2+x+1 nor
x^3+x+1 can be divisible by 5; x^4+x+1 is sometimes divisible
by 15, and x^5+x+1 is sometimes divisible by 105.


On a somewhat related topic, note that any abundant number of
the form n^2+n+41 must have at least 100,000 prime factors.
And although it is within our computational capability to
construct an abundant number of this form, and to prove that
it is abundant, it's probably out of our reach to actually
find its complete prime factorization.  :)