OFFSET
1,2
COMMENTS
For n > 1, n! cannot be a perfect square. Proof: All exponents of the prime factors of a square are even. But in the factorization of n! at least one of the primes will appear only once due to Bertrand's Postulate which says there is always a prime between m and 2m.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..100
EXAMPLE
a(9)=67 because 9!+729 = 363609 = 3^4*67^2 is a square with largest prime factor of 67.
MATHEMATICA
Join[{0}, Table[FactorInteger[(Floor[Sqrt[n!]]+1)^2][[-1, 1]], {n, 2, 30}]] (* Harvey P. Dale, Jan 04 2012 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jason Earls, Jun 06 2003
STATUS
approved