%I A082022
%S A082022 1,4,18,576,1200,518400,1587600,180633600,1646023680,13168189440000,
%T A082022 461039040000,229442532802560000,86553098311680000,3753113311877529600,
%U A082022 834966920275488000000
%N A082022 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 A082022 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 A082022 Prod(k=1...n, lcm(k, n+1-k)).
%e A082022 1 2 3 4 5...
%e A082022 2 2 6 4 10...
%e A082022 3 6 3 12 15...
%e A082022 4 4 12 4 20...
%e A082022 5 10 15 20 5...
%e A082022 ...
%e A082022 The same array in triangular form is
%e A082022 1
%e A082022 2 2
%e A082022 3 2 3
%e A082022 4 6 6 4
%e A082022 5 4 3 4 5
%e A082022 ...
%e A082022 Sequence contains the product of the terms of the n-th row.
%o A082022 (PARI) for(n=1,20,p=1:for(k=1,n,p=p*lcm(k,n+1-k)):print1(p","))
%Y A082022 Cf. A006580, A003990, A082292.
%Y A082022 Equals A001044(n) / A051190(n+1).
%Y A082022 Sequence in context: A145660 A156522 A086400 this_sequence A114574 A156491 
               A069874
%Y A082022 Adjacent sequences: A082019 A082020 A082021 this_sequence A082023 A082024 
               A082025
%K A082022 nonn
%O A082022 1,2
%A A082022 Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Apr 06 2003
%E A082022 Corrected and extended by Ralf Stephan (ralf(AT)ark.in-berlin.de), Apr 
               08 2003