A082022 - OEIS

%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