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A157612
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Number of factorizations of n! into distinct factors.
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9
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1, 1, 1, 2, 5, 16, 57, 253, 1060, 5285, 28762, 191263, 1052276, 8028450, 56576192, 424900240, 2584010916, 24952953943, 178322999025, 1886474434192, 15307571683248, 143131274598786, 1423606577935925, 17668243239613767, 137205093278725072, 1399239022852163764, 15774656316828338767
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OFFSET
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0,4
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COMMENTS
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The number of factorizations of (n+1)! into k distinct factors can be arranged into the following triangle:
2! 1;
3! 1, 1;
4! 1, 3, 1;
5! 1, 7, 7, 1;
...
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LINKS
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FORMULA
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EXAMPLE
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3! = 6 = 2*3.
a(3) = 2 because there are 2 factorizations of 3!.
4! = 24 = 2*12 = 3*8 = 4*6 = 2*3*4.
a(4) = 5 because there are 5 factorizations of 4!.
5! = 120 (1)
5! = 2*60 = 3*40 = 4*30 = 5*24 = 6*20 = 8*15 = 10*12 (7)
5! = 2*3*20 = 2*4*15 = 2*5*12 = 2*6*10 = 3*4*10 = 3*5*8 = 4*5*6 (7)
5! = 2*3*4*5 (1)
a(5) = 16 because there are 16 factorizations of 5!.
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MAPLE
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with(numtheory):
b:= proc(n, k) option remember;
`if`(n>k, 0, 1) +`if`(isprime(n), 0,
add(`if`(d>k, 0, b(n/d, d-1)), d=divisors(n) minus {1, n}))
end:
a:= n-> b(n!$2):
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MATHEMATICA
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b[n_, k_] := b[n, k] = If[n>k, 0, 1] + If[PrimeQ[n], 0, Sum[If[d>k, 0, b[n/d, d-1]], {d, Divisors[n] ~Complement~ {1, n}}]];
a[n_] := b[n!, n!];
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PROG
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a(n)={if(n<=1, 1, count(factor(n!)[, 2]))} \\ Andrew Howroyd, Feb 01 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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