A129365 -id:A129365 - cp's OEIS frontend

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

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

A092287 a(n) = Product_{j=1..n} Product_{k=1..n} gcd(j,k).

Original entry on oeis.org

1, 1, 2, 6, 96, 480, 414720, 2903040, 5945425920, 4334215495680, 277389791723520000, 3051287708958720000, 437332621360674939863040000, 5685324077688774218219520000, 15974941971638268369709427589120000, 982608696336737613503095822614528000000000

Offset: 0

N. J. A. Sloane, based on a suggestion from Leroy Quet, Feb 03 2004

nonn

Conjecture: Let p be a prime and let ordp(n,p) denote the exponent of the highest power of p that divides n. For example, ordp(48,2)=4, since 48=3*(2^4). Then we conjecture that the prime factorization of a(n) is given by the formula: ordp(a(n),p) = (floor(n/p))^2 + (floor(n/p^2))^2 + (floor(n/p^3))^2 + .... Compare this to the de Polignac-Legendre formula for the prime factorization of n!: ordp(n!,p) = floor(n/p) + floor(n/p^2) + floor(n/p^3) + .... This suggests that a(n) can be considered as generalization of n!. See A129453 for the analog for a(n) of Pascal's triangle. See A129454 for the sequence defined as a triple product of gcd(i,j,k). - Peter Bala, Apr 16 2007
The conjecture is correct. - Charles R Greathouse IV, Apr 02 2013
a(n)/a(n-1) = n, n >= 1, if and only if n is noncomposite, otherwise a(n)/a(n-1) = n * f^2, f > 1. - Daniel Forgues, Apr 07 2013
Conjecture: For a product over a rectangle, f(n,m) = Product_{j=1..n} Product_{k=1..m} gcd(j,k), a factorization similar to the one given above for the square case takes place: ordp(f(n,m),p) = floor(n/p)*floor(m/p) + floor(n/p^2)*floor(m/p^2) + .... By way of directly computing the values of f(n,m), it can be verified that the conjecture holds at least for all 1 <= m <= n <= 200. - Andrey Kaydalov, Mar 11 2019

T. D. Noe, Table of n, a(n) for n = 0..67

Cf. A003989, A018806, A051190, A090494, A129365, A129439, A129453, A129454, A129455, A224479, A224497.

[n eq 0 select 1 else (&*[(&*[GCD(j,k): k in [1..n]]): j in [1..n]]): n in [0..30]]; // G. C. Greubel, Feb 07 2024

f := n->mul(mul(igcd(j,k),k=1..n),j=1..n);

a[0] = 1; a[n_] := a[n] = n*Product[GCD[k, n], {k, 1, n-1}]^2*a[n-1]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Apr 16 2013, after Daniel Forgues *)

h(n,p)=if(nCharles R Greathouse IV, Apr 02 2013

def A092287(n):
    R = 1
    for p in primes(n+1) :
        s = 0; r = n
        while r > 0 :
            r = r//p
            s += r*r
        R *= p^s
    return R
[A092287(i) for i in (0..15)]  # Peter Luschny, Apr 10 2013

Also a(n) = Product_{k=1..n} Product_{j=1..n} lcm(1..floor(min(n/k, n/j))).
From Daniel Forgues, Apr 08 2013: (Start)
Recurrence: a(0) := 1; for n > 0: a(n) := n * (Product_{j=1..n-1} gcd(n,j))^2 * a(n-1) = n * A051190(n)^2 * a(n-1).
Formula for n >= 0: a(n) = n! * (Product_{j=1..n} Product_{k=1..j-1} gcd(j,k))^2. (End)
a(n) = n! * A224479(n)^2 (the last formula above).
a(n) = n$ * A224497(n)^4, n$ the swinging factorial A056040(n). - Peter Luschny, Apr 10 2013

Recurrence formula corrected by Daniel Forgues, Apr 07 2013

A129364 a(n) = Product_{k = 1..n} A066841(k).

Original entry on oeis.org

1, 2, 6, 96, 480, 207360, 1451520, 2972712960, 722369249280, 5778953994240000, 63568493936640000, 9111096278347394580480000, 118444251618516129546240000, 10400352846118664303196241920000

Offset: 1

Peter Bala, Apr 11 2007

Conjecture: a(n) divides A092287(n) for all n - see comments in A129365.

Cf. A007955, A066841, A092143, A092287, A129365.

Table[Product[Floor[n/k]!^k, {k, 1, n}], {n, 1, 15}] (* Vaclav Kotesovec, Jun 24 2021 *)
Table[Product[k^(Floor[n/k]*(1 + Floor[n/k])/2), {k, 1, n}], {n, 1, 15}] (* Vaclav Kotesovec, Jun 24 2021 *)

a(n) = prod(k=1, n, k^((n\k) * (1 + n\k) \ 2)); \\ Daniel Suteu, Sep 12 2018

a(n) = Product_{k = 1..n} Product_{d|k} d^(k/d).
a(n) = Product_{k = 1..n} ((floor(n/k))!)^k.
a(n) = exp(Sum_{k = 1..n} log(k)/2 * floor(n/k) * floor(1 + n/k)). - Daniel Suteu, Sep 12 2018
log(a(n)) ~ c * n^2, where c = -zeta'(2)/2 = A073002/2 = 0.468774... - Vaclav Kotesovec, Jun 24 2021

A090494 Product_{j=1..n} Product_{k=1..n} lcm(j,k).

Original entry on oeis.org

1, 1, 8, 7776, 1146617856, 1289945088000000000, 46798828032806092800000000000, 2350577043461005964030008507760640000000000000, 8206262459636402163263383676462776103575725539328000000000000000, 2746781358330240881921653545637784861521126603512175621574459373964492800000000000000000

Offset: 0

N. J. A. Sloane, Feb 03 2004

nonn

Cf. A018806.

Cf. A092143, A092287, A129365, A129454.

f := n->mul(mul(lcm(j,k),k=1..n),j=1..n);

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). Then the prime factorization of a(n) appears to be given by the formula ordp(a(n),p)= sum_{k >= 1} [(2*(p^k)-1)*floor((n/(p^k)))^2] + 2*sum_{k >= 1} [floor(n/(p^k))*mod(n,p^k)], for each prime p. See the comments sections of A092143, A092287, A129365 and A129454 for similar conjectural prime factorizations. - Peter Bala, Apr 23 2007

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

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A092287 a(n) = Product_{j=1..n} Product_{k=1..n} gcd(j,k).

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A129364 a(n) = Product_{k = 1..n} A066841(k).

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A090494 Product_{j=1..n} Product_{k=1..n} lcm(j,k).

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