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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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%I #21 May 03 2022 21:16:20

%S 1,4,36,2304,57600,74649600,3657830400,14982473318400,

%T 10922223049113600,109222230491136000000,13215889889427456000000,

%U 39462435755592152776704000000,6669151642695073819262976000000,256202129505773955840806486016000000

%N a(n) = Product_{k=1..n} k^d(k), where d(k) = number of divisors of k.

%C a(n) = a(n-1)*A062758(n).

%C a(n) = Product_{k=1..n} k^floor(n/k) * (floor(n/k))!.

%H Michael S. Branicky, <a href="/A175493/b175493.txt">Table of n, a(n) for n = 1..117</a>

%t f[n_] := Product[ k^DivisorSigma[0, k], {k, n}]; Array[f, 15] (* _Robert G. Wilson v_, Jun 11 2010 *)

%o (Python)

%o from sympy import divisor_count

%o from itertools import count, islice

%o def agen():

%o an = 1

%o for k in count(2):

%o yield an

%o an *= k**divisor_count(k)

%o print(list(islice(agen(), 14))) # _Michael S. Branicky_, May 03 2022

%o (PARI) a(n) = prod(k=1, n, k^numdiv(k)); \\ _Michel Marcus_, May 03 2022

%Y Cf. A062758.

%Y Cf. A174939 (sum instead of product).

%K nonn

%O 1,2

%A _Leroy Quet_, May 30 2010

%E a(6) onwards from _Robert G. Wilson v_ and _Jon E. Schoenfield_, Jun 11 2010

%E a(14) and beyond from _Michael S. Branicky_, May 03 2022