OFFSET
0,4
COMMENTS
It is possible to prove that, if gcd(k,a,b) = 1, then k^a + a^b + b^k = n! can be solved only if a = b = 1, thus k = n! - 2 for every n > 2.
LINKS
M Cipu, F. Luca and M. Mignotte, Solutions of the diophantine equation x^y+y^z+z^x=n!, Glasgow Mathematical Journal, 50(2008), 217-232.
FORMULA
E.g.f.: 1/(1-x) - 2*exp(x). - Alois P. Heinz, Sep 10 2015
MAPLE
MATHEMATICA
Table[n! - 2, {n, 20}] (* Wesley Ivan Hurt, Aug 13 2015 *)
PROG
(Magma) [Factorial(n)-2 : n in [1..20]]; // Wesley Ivan Hurt, Aug 13 2015
(PARI) a(n)=n!-2 \\ Charles R Greathouse IV, Aug 28 2015
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Marco Ripà, Aug 11 2015
EXTENSIONS
a(0)-a(1) corrected by David A. Corneth, Sep 10 2015
STATUS
approved