A092143 Cumulative product of all divisors of 1..n.

1, 2, 6, 48, 240, 8640, 60480, 3870720, 104509440, 10450944000, 114960384000, 198651543552000, 2582470066176000, 506164132970496000, 113886929918361600000, 116620216236402278400000, 1982543676018838732800000, 11562194718541867489689600000, 219681699652295482304102400000

Offset: 1

Author

Jon Perry, Mar 31 2004

Keywords

• nonn

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. - Peter Bala, Apr 15 2007

Examples

			a(6) = 1*2*3*2*4*5*2*3*6 = 8640.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..190
• Angelo B. Mingarelli, Abstract factorials, arXiv:0705.4299 [math.NT], 2007-2012.

Crossrefs

Cf. A006218, A010786, A092144, A117871.

Cf. A129364, A129365, A129439, A345466.

Cf. A000005, A000178, A007955, A175493, A280714.

Programs

Magma

[(&*[j^Floor(n/j): j in [1..n]]): n in [1..30]]; // G. C. Greubel, Feb 05 2024

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 *)

PARI

my(z=1); for(i=1,25, fordiv(i,j,z*=j); print1(z, ", "))

SageMath

[product(j^(n//j) for j in range(1,n+1)) for n in range(1,31)] # G. C. Greubel, Feb 05 2024

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