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%I #48 Jun 10 2019 01:11:29
%S 1,1,3,1,9,27,15,18,288,288,420,464,1856,10080,46848,210240,400320,
%T 652848,3991680,27528402,32659200,163296000,1143463200,1305467240,
%U 6840489600,9453465438,337082683248,163425485250,8376514506360,8440230839040,5088099594240
%N Least k such that n! + k^2 is a square.
%C 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
%C 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
%H Sudipta Mallick, <a href="/A038202/b038202.txt">Table of n, a(n) for n = 4..58</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BrocardsProblem.html">Brocard's Problem</a>
%t Table[f=n!/4; x=Max[Select[Divisors[f], #<=Sqrt[f]&]]; f/x-x, {n, 4, 20}] (* _T. D. Noe_, Nov 02 2004 *)
%o (PARI) a(n) = my(k=0); while(!issquare(n!+k^2), k++); k; \\ _Michel Marcus_, Sep 16 2018
%Y Cf. A038548 (number of divisors of n that are at most sqrt(n)), A068869.
%Y Cf. A181892, A181896.
%K nonn
%O 4,3
%A _David W. Wilson_
%E a(30)-a(34) from _Jon E. Schoenfield_, Sep 15 2018
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