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A175493
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a(n) = Product_{k=1..n} k^d(k), where d(k) = number of divisors of k.
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2
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1, 4, 36, 2304, 57600, 74649600, 3657830400, 14982473318400, 10922223049113600, 109222230491136000000, 13215889889427456000000, 39462435755592152776704000000, 6669151642695073819262976000000, 256202129505773955840806486016000000
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OFFSET
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1,2
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COMMENTS
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a(n) = Product_{k=1..n} k^floor(n/k) * (floor(n/k))!.
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LINKS
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MATHEMATICA
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f[n_] := Product[ k^DivisorSigma[0, k], {k, n}]; Array[f, 15] (* Robert G. Wilson v, Jun 11 2010 *)
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PROG
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(Python)
from sympy import divisor_count
from itertools import count, islice
def agen():
an = 1
for k in count(2):
yield an
an *= k**divisor_count(k)
(PARI) a(n) = prod(k=1, n, k^numdiv(k)); \\ Michel Marcus, May 03 2022
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CROSSREFS
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Cf. A174939 (sum instead of product).
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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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