OFFSET
4,3
COMMENTS
Let f = n!/4 and let x be the largest divisor of f such that x < sqrt(f). Then a(n) = f/x - x. The greatest k such that n! + k^2 is a square is f-1. The number of k for which n! + k^2 is a square is A038548(n). - T. D. Noe, Nov 02 2004
For greatest k such that n! + k^2 is a square see A181892; for numbers x such that n! + k^2 = x^2 see A181896. - Artur Jasinski, Mar 31 2012
LINKS
Sudipta Mallick, Table of n, a(n) for n = 4..58
Eric Weisstein's World of Mathematics, Brocard's Problem
MATHEMATICA
Table[f=n!/4; x=Max[Select[Divisors[f], #<=Sqrt[f]&]]; f/x-x, {n, 4, 20}] (* T. D. Noe, Nov 02 2004 *)
PROG
(PARI) a(n) = my(k=0); while(!issquare(n!+k^2), k++); k; \\ Michel Marcus, Sep 16 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
a(30)-a(34) from Jon E. Schoenfield, Sep 15 2018
STATUS
approved