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A343743
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a(n) is the largest base in which the order of the Monster group has (47 - n) zeros; alternatively, radicals of maximal powers dividing the order of the Monster group.
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1
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2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 12, 12, 12, 12, 12, 24, 24, 24, 24, 48, 144, 1440, 1440, 2880, 120960, 1451520, 87091200, 1902071808000, 15184923989114880000, 808017424794512875886459904961710757005754368000000000
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OFFSET
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1,1
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COMMENTS
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Let z be a specified minimum number of zeros in the order of the Monster group; here z is a natural number, 1 <= z <= 46, with z = (47 - n). Then the largest base in which the order of the Monster group has at least z zeros is:
Product_{k=1..20} prime(k)^floor(A051161(k)/z).
When z = 1 this is the order of the Monster group.
Every term in this sequence except the last is a number of least prime signature (A025487).
In the following table, when the order of the Monster group has exactly z zeros, it also has s significant digits, and d = s + z total digits.
z s d
-- --- ---
46 134 180
23 67 90
20 30 50
15 25 40
11 22 33
10 15 25
9 9 18
7 9 16
6 5 11
5 4 9
4 3 7
3 2 5
2 1 3
1 1 2
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REFERENCES
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J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker and R. A. Wilson, ATLAS of Finite Groups. Oxford Univ. Press, 1985.
J. H. Conway, N. J. A. Sloane, Sphere Packings, Lattices, and Groups. Springer, 3rd ed., 1999.
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LINKS
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EXAMPLE
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a(27) = the largest base in which the order of the Monster group has at least (47 - 27) = 20 zeros. This is 2^(floor(46/20)) * 3^(floor(20/20)) = 2^2 * 3 = 12; the remaining terms in the product have exponent 0.
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MATHEMATICA
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f = FactorInteger[MonsterGroupM[] // GroupOrder]; Table[Times @@ ((First[#]^Floor[Last[#]/z]) & /@ f), {z, Max[f[[;; , 2]]], 1, -1}] (* Amiram Eldar, Jul 19 2021 *)
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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STATUS
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approved
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