A358070 Largest order of element in direct product S_n * S_n where S_n is the symmetric group.

1, 1, 2, 6, 12, 30, 30, 84, 120, 210, 420, 420, 840, 1260, 2310, 4620, 5460, 5460, 13860, 13860, 27720, 32760, 60060, 60060, 120120, 180180, 180180, 360360, 360360, 510510, 1021020, 1141140, 2042040, 3063060, 3423420, 6126120, 6846840, 6846840, 8953560, 12252240

Offset: 0

Author

Jack Zhang, Oct 29 2022

Keywords

• nonn

Comments

Let (P,Q) be two partitions of n and lcm(P) be the LCM of all parts of P, then a(n) = max( lcm(lcm(P), lcm(Q)) ) where the maximum is taken among all pairs (P,Q). - Joerg Arndt, Dec 04 2022

Examples

			From _Joerg Arndt_, Dec 04 2022: (Start)
The 15 partitions of 7 are the following:
[ #]  [ partition ]   lcm( parts )
[ 1]  [ 1 1 1 1 1 1 1 ]   1
[ 2]  [ 1 1 1 1 1 2 ]   2
[ 3]  [ 1 1 1 1 3 ]   3
[ 4]  [ 1 1 1 2 2 ]   2
[ 5]  [ 1 1 1 4 ]   4
[ 6]  [ 1 1 2 3 ]   6
[ 7]  [ 1 1 5 ]   5
[ 8]  [ 1 2 2 2 ]   2
[ 9]  [ 1 2 4 ]   4
[10]  [ 1 3 3 ]   3
[11]  [ 1 6 ]   6
[12]  [ 2 2 3 ]   6
[13]  [ 2 5 ]  10
[14]  [ 3 4 ]  12
[15]  [ 7 ]   7
The maximum value attained is 7 * 12, so a(7) = 84.
(End)
		

Crossrefs

Cf. A000793 (largest order of element in S_n).

Cf. A063183.

Programs

Python3

x=[{1},{1}]
for i in range(2,40):
    u=[]
    for j in range(1,i):
        u.extend([k*j//math.gcd(k,j) for k in x[i-j]])
    x.append(set(u))
xx=[set([i*j//math.gcd(i,j) for i in t for j in t]) for t in x]
print([max(i) for i in xx][2:])