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.

1, 1, 4, 18, 576, 1200, 518400, 1587600, 180633600, 1646023680, 13168189440000, 461039040000, 229442532802560000, 86553098311680000, 3753113311877529600, 834966920275488000000, 437763136697395052544000000, 677906091957878784000000, 40990389067797283140009984000000

Offset

0,3

Comments

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

Links

Paolo Xausa, Table of n, a(n) for n = 0..250

Formula

a(n) = Product_{k=1..n} lcm(k, n+1-k).

Example

1 2 3 4 5...
2 2 6 4 10...
3 6 3 12 15...
4 4 12 4 20...
5 10 15 20 5...
...
The same array in triangular form is
1
2 2
3 2 3
4 6 6 4
5 4 3 4 5
...
Sequence contains the product of the terms of the n-th row.

Maple

a:= n-> mul(ilcm(i, 1+n-i), i=1..n):
seq(a(n), n=0..18);  # Alois P. Heinz, Nov 23 2025

Mathematica

A082022[n_] := n!^2/Product[GCD[k, n + 1], {k, n}];
Array[A082022, 20, 0] (* Paolo Xausa, Nov 24 2025 *)

Prog

(PARI) for(n=0, 20, p=1; for(k=1, n, p=p*lcm(k, n+1-k)); print1(p, ", "))

Crossrefs

Cf. A006580, A003990, A082292.

Equals A001044(n) / A051190(n+1).

Sequence in context: A214240 A156522 A086400 * A354303 A114574 A383772

Adjacent sequences: A082019 A082020 A082021 * A082023 A082024 A082025

Keyword

nonn

Author

Amarnath Murthy, Apr 06 2003

Extensions

Corrected and extended by Ralf Stephan, Apr 08 2003

a(0)=1 prepended by Alois P. Heinz, Nov 23 2025

Status

approved