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

0,3

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

Links

Table of n, a(n) for n=0..39.

Example

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)

Prog

(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:])

Crossrefs

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

Cf. A063183.

Sequence in context: A195166 A225646 A225627 * A237704 A181826 A143176

Adjacent sequences: A358067 A358068 A358069 * A358071 A358072 A358073

Keyword

nonn

Author

Jack Zhang, Oct 29 2022

Status

approved