A129364 -id:A129364 - cp's OEIS frontend

A129365 a(n) = A092287(n)/A129364(n).

1, 1, 1, 1, 1, 2, 2, 2, 6, 48, 48, 48, 48, 1536, 207360, 207360, 207360, 1105920, 1105920, 17694720, 30098718720, 15410543984640, 15410543984640, 481579499520, 60197437440000, 123284351877120000, 29958097506140160000

Offset: 1

Peter Bala, Apr 13 2007

Conjectures:
A) a(n) is always an integer.
B) If p is a prime then p|a(n) if and only if p <= n/3. 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). The precise decomposition of a(n) into primes would follow from the following two conjectures:
C) For each positive integer n and prime p, ordp(a(n*p),p) = ordp(a(n*p+1),p) = ordp(a(n*p+2),p) = . . . = ordp(a(n*p+p-1),p).
D) Let b(n) = A004125(n). Then ordp(a(n*p),p) = b(n) + b(floor(n/p)) + b(floor(n/p^2)) + b(floor(n/p^3)) + .... This is reminiscent of de Polignac's formula (also due to Legendre) for the prime factorization of n! (see the link).

Wikipedia, De Polignac's formula.

Cf. A004125, A092287, A129364.

a(n) = ( Product_{j = 1..n} Product_{k = 1..n} gcd(j,k) ) / ( Product_{j = 1..n} Product_{d|j} d^(j/d) ).

a(n) = ( Product_{j = 1..n} Product_{k = 1..n} gcd(j,k) ) / ( Product_{k = 1..n} (floor(n/k)!)^k ).

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

Original entry on oeis.org

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

Offset: 1

Jon Perry, Mar 31 2004

nonn

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

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

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

Cf. A006218, A010786, A092144, A117871.
Cf. A129364, A129365, A129439, A345466.
Cf. A000005, A000178, A007955, A175493, A280714.

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

seq(sqrt(mul(k^numtheory[tau](k), k=1..n)), n=1..40); # Ridouane Oudra, Oct 31 2024

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 *)

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

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

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

A345726 a(n) = Product_{k=1..n} k^(floor(n/k)^2).

Original entry on oeis.org

1, 2, 6, 192, 960, 4976640, 34836480, 2283043553280, 4993016251023360, 3195530400654950400000, 35150834407204454400000, 417877827219530751882239882035200000, 5432411753853899774469118466457600000, 213700126654516647665669790727613605478400000

Offset: 1

Vaclav Kotesovec, Jun 25 2021

nonn

G. C. Greubel, Table of n, a(n) for n = 1..55

Cf. A073002, A092143, A129364.

[(&*[j^(Floor(n/j))^2: j in [1..n]]): n in [1..30]]; // G. C. Greubel, Feb 05 2024

Table[Product[k^(Floor[n/k]^2), {k, 1, n}], {n, 1, 15}]

a(n) = prod(k=1, n, k^((n\k)^2)); \\ Michel Marcus, Jun 26 2021

[product(j^((n//j)^2) for j in range(1,n+1)) for n in range(1,31)] # G. C. Greubel, Feb 05 2024

log(a(n)) ~ c * n^2, where c = -zeta'(2) = A073002.

cp's OEIS Frontend

A129365 a(n) = A092287(n)/A129364(n).

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

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A345726 a(n) = Product_{k=1..n} k^(floor(n/k)^2).

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Magma

Mathematica

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