Diophantine equation with factorial

[all at abouthugo.de]

Don McDonald <parabola at paradise.net.nz> schrieb am 22.06.2003, 15:22:42:
> In message  you write:
> 
> > 
> > SeqFans,
> > 
> > a few days ago the following was posted to
> > NG sci.math.research. No reply so far.
> > At least there don't seem too many solutions....
> > Any ideas?
> > 
> > Hugo 
> 
> I tested up to 20! = 2.4E18.
> But my Basic is only good for 4.5E15. (by trial.)
> 
> Tip:
> start at x = (n! /2) ^(1/3).
> x increases by a factor of cubert 2.
> 
> y decreases rapidly.
> This greatly reduces the number of trials.
> 
> The next (and better) idea is try x^3+y^3 mod 21.
> Only 200 easy trials needed.
> 
> don.mcdonald
> 23.06.03  01:12
> 
> ..Calc.Profile.eisintegsq.Seqfan.x3+y3=nfac
> > >
> > > -------- Original Message --------
> > > Betreff: Diophantine equation with factorial
> > > Datum: Thu, 19 Jun 2003 15:26:48 GMT
> > > Von: Alex Liu 
> > > Foren: sci.math.research
> > > 
> > > It is clear that (x,y,n)=(1,1,2) is [a] positive integer solution to
> > > the equation x^3+y^3=(n!). Any other positive integer solution to this
> > > equation?
> >

My original hope was that I could simply do a
lookup, but I couldn't find the following (just
submitted):

%S A000001 1 2 6 3 5
%N A000001 Least number of positive cubes needed to represent n!.
%F A000001 a(n)=A002376(n!).
%e A000001 a(4)=3 because 4!=24=2^3+2^3+2^3.
%Y A000001 Cf. A000142, A002376, A002325, A003072, A003327, A003328,
A003329.
%O A000001 1
%K A000001 ,more,nonn,
%A A000001 Hugo Pfoertner (hugo at pfoertner.org), Jun 22 2003

Many more terms needed; a nice task taking into
account the 3 line convention.

Hugo