Search: id:A038202
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%I A038202
%S A038202 1,1,3,1,9,27,15,18,288,288,420,464,1856,10080,46848,210240,400320,
%T A038202 652848,3991680,27528402,32659200,163296000,1143463200,1305467240,
%U A038202 6840489600,9453465438
%N A038202 Least k such that n!+k^2 is a square.
%C A038202 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(f). 
               - T. D. Noe (noe(AT)sspectra.com), Nov 02 2004
%H A038202 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               BrocardsProblem.html">Link to a section of The World of Mathematics.</
               a>
%t A038202 Table[f=n!/4; x=Max[Select[Divisors[f], #<=Sqrt[f]&]]; f/x-x, {n, 4, 
               20}] (T. D. Noe)
%Y A038202 Cf. A038548 (number of divisors of n that are at most sqrt(n)).
%Y A038202 Sequence in context: A160568 A157403 A105951 this_sequence A128415 A090479 
               A141903
%Y A038202 Adjacent sequences: A038199 A038200 A038201 this_sequence A038203 A038204 
               A038205
%K A038202 nonn
%O A038202 4,3
%A A038202 David W. Wilson (davidwwilson(AT)comcast.net)

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