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A092143
Cumulative product of all divisors of 1..n.
16
1, 2, 6, 48, 240, 8640, 60480, 3870720, 104509440, 10450944000, 114960384000, 198651543552000, 2582470066176000, 506164132970496000, 113886929918361600000, 116620216236402278400000, 1982543676018838732800000, 11562194718541867489689600000, 219681699652295482304102400000
OFFSET
1,2
COMMENTS
Let p be a prime and let ordp(n,p) denote the exponent of the largest power of p which divides n. For example, ordp(48,2) = 4 since 48 = 3*(2^4). Let b(n) = A006218(n) = Sum_{k = 1..n} floor(n/k). The prime factorization of a(n) appears to be given by the following conjectural formula: ordp(a(n),p) = b(floor(n/p)) + b(floor(n/p^2)) + b(floor(n/p^3)) + ... . Compare with the comments in A129365. [added 13 May 2026: for a proof of this conjecture by Grok AI see the Bala link.] - Peter Bala, Apr 15 2007
LINKS
Angelo B. Mingarelli, Abstract factorials, arXiv:0705.4299 [math.NT], 2007-2012.
FORMULA
a(n) = Product_{k=1..n} {floor(n/k)}!. This formula is due to Sebastian Martin Ruiz. - Peter Bala, Apr 15 2007; Formula corrected by R. J. Mathar, May 06 2008
Sum_{n>=1} 1/a(n) = A117871. - Amiram Eldar, Nov 17 2020
log(a(n)) ~ n * log(n)^2 / 2. - Vaclav Kotesovec, Jun 20 2021
a(n) = Product_{k=1..n} k^floor(n/k). - Vaclav Kotesovec, Jun 24 2021
From Ridouane Oudra, Oct 31 2024: (Start)
a(n) = Product_{k=1..n} A007955(k).
a(n) = Product_{k=1..n} k^(tau(k)/2).
a(n) = sqrt(A175493(n)). (End)
a(n) = A000178(n)/A280714(n). - Amiram Eldar, Aug 16 2025
EXAMPLE
a(6) = 1*2*3*2*4*5*2*3*6 = 8640.
MAPLE
seq(sqrt(mul(k^numtheory[tau](k), k=1..n)), n=1..40); # Ridouane Oudra, Oct 31 2024
MATHEMATICA
Reap[For[n = k = 1, k <= 25, k++, Do[n = n*d, {d, Divisors[k]}]; Sow[n]]][[2, 1]] (* Jean-François Alcover, Oct 30 2012 *)
Table[Product[k^Floor[n/k], {k, 1, n}], {n, 1, 25}] (* Vaclav Kotesovec, Jun 24 2021 *)
FoldList[Times, Times @@@ Divisors[Range[25]]] (* Paolo Xausa, Nov 06 2024 *)
PROG
(PARI) my(z=1); for(i=1, 25, fordiv(i, j, z*=j); print1(z, ", "))
(Magma) [(&*[j^Floor(n/j): j in [1..n]]): n in [1..30]]; // G. C. Greubel, Feb 05 2024
(SageMath) [product(j^(n//j) for j in range(1, n+1)) for n in range(1, 31)] # G. C. Greubel, Feb 05 2024
(Python)
from sympy import factorial
def A092143(n):
c, j = 1, 1
while j <= n:
c *= factorial(k:=n//j)**(-j+(j:=n//k+1))
return c # Chai Wah Wu, Jan 29 2026
KEYWORD
nonn,easy,changed
AUTHOR
Jon Perry, Mar 31 2004
STATUS
approved