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

%I A358070 #22 Jan 04 2023 18:47:13
%S A358070 1,1,2,6,12,30,30,84,120,210,420,420,840,1260,2310,4620,5460,5460,
%T A358070 13860,13860,27720,32760,60060,60060,120120,180180,180180,360360,
%U A358070 360360,510510,1021020,1141140,2042040,3063060,3423420,6126120,6846840,6846840,8953560,12252240
%N A358070 Largest order of element in direct product S_n * S_n where S_n is the symmetric group.
%C A358070 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
%e A358070 From _Joerg Arndt_, Dec 04 2022: (Start)
%e A358070 The 15 partitions of 7 are the following:
%e A358070 [ #]  [ partition ]   lcm( parts )
%e A358070 [ 1]  [ 1 1 1 1 1 1 1 ]   1
%e A358070 [ 2]  [ 1 1 1 1 1 2 ]   2
%e A358070 [ 3]  [ 1 1 1 1 3 ]   3
%e A358070 [ 4]  [ 1 1 1 2 2 ]   2
%e A358070 [ 5]  [ 1 1 1 4 ]   4
%e A358070 [ 6]  [ 1 1 2 3 ]   6
%e A358070 [ 7]  [ 1 1 5 ]   5
%e A358070 [ 8]  [ 1 2 2 2 ]   2
%e A358070 [ 9]  [ 1 2 4 ]   4
%e A358070 [10]  [ 1 3 3 ]   3
%e A358070 [11]  [ 1 6 ]   6
%e A358070 [12]  [ 2 2 3 ]   6
%e A358070 [13]  [ 2 5 ]  10
%e A358070 [14]  [ 3 4 ]  12
%e A358070 [15]  [ 7 ]   7
%e A358070 The maximum value attained is 7 * 12, so a(7) = 84.
%e A358070 (End)
%o A358070 (Python3)
%o A358070 x=[{1},{1}]
%o A358070 for i in range(2,40):
%o A358070     u=[]
%o A358070     for j in range(1,i):
%o A358070         u.extend([k*j//math.gcd(k,j) for k in x[i-j]])
%o A358070     x.append(set(u))
%o A358070 xx=[set([i*j//math.gcd(i,j) for i in t for j in t]) for t in x]
%o A358070 print([max(i) for i in xx][2:])
%Y A358070 Cf. A000793 (largest order of element in S_n).
%Y A358070 Cf. A063183.
%K A358070 nonn
%O A358070 0,3
%A A358070 _Jack Zhang_, Oct 29 2022