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%I #6 Apr 30 2014 01:27:34
%S 1,4,18,576,1200,518400,1587600,180633600,1646023680,13168189440000,
%T 461039040000,229442532802560000,86553098311680000,
%U 3753113311877529600,834966920275488000000
%N In the following square array a(i,j) = Least Common Multiple of i and j. Sequence contains the product of the terms of the n-th antidiagonal.
%C If n is even and n+1 is prime, a(n) = n^2 * (n-1)!^2. If n is odd and >3, 2*(n+1)*a(n) is a perfect square, the root of which has the factor 1/2*n*(n-1)*((n-1)/2)!. This was proved by Lawrence Sze. - _Ralf Stephan_, Nov 16 2004
%F Prod(k=1...n, lcm(k, n+1-k)).
%e 1 2 3 4 5...
%e 2 2 6 4 10...
%e 3 6 3 12 15...
%e 4 4 12 4 20...
%e 5 10 15 20 5...
%e ...
%e The same array in triangular form is
%e 1
%e 2 2
%e 3 2 3
%e 4 6 6 4
%e 5 4 3 4 5
%e ...
%e Sequence contains the product of the terms of the n-th row.
%o (PARI) for(n=1,20,p=1:for(k=1,n,p=p*lcm(k,n+1-k)):print1(p","))
%Y Cf. A006580, A003990, A082292.
%Y Equals A001044(n) / A051190(n+1).
%K nonn
%O 1,2
%A _Amarnath Murthy_, Apr 06 2003
%E Corrected and extended by _Ralf Stephan_, Apr 08 2003
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