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A358070
Largest order of element in direct product S_n * S_n where S_n is the symmetric group.
0
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
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
KEYWORD
nonn
AUTHOR
Jack Zhang, Oct 29 2022
STATUS
approved