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Largest divisor of n!/4 which is less than sqrt(n!)/2.
0

%I #15 Jun 22 2019 10:49:21

%S 2,5,12,35,96,288,945,3150,10800,39312,147420,571536,2286144,9424800,

%T 39984000,174283200,779688000,3573570000,16761064320,80379048750,

%U 393826406400,1969132032000,10040487256800,52174220175000,276080056560000,1486750296281250

%N Largest divisor of n!/4 which is less than sqrt(n!)/2.

%C Comment from A038202: Let f=n!/4 and let a(n) be the largest divisor of f such that a(n) < sqrt(f). Then A038202(n) = f/a(n) - a(n). 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(f). - _T. D. Noe_, Nov 02 2004

%t Table[f = n!/4; Select[Divisors[f], # <= Sqrt[f] &][[-1]], {n, 4, 20}]

%Y Cf. A038202, A138196, A139151, A181892, A181893.

%K nonn

%O 4,1

%A _Artur Jasinski_, Mar 31 2012