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

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

Offset

0,3

Comments

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

Links

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

Formula

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

Maple

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

Mathematica

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 *)

Prog

(PARI) h(n, p)=if(n<p, 0, n\=p; h(n, p)+n^2)
a(n)=prod(i=1, primepi(n), my(p=prime(i)); p^h(n, p)) \\ Charles R Greathouse IV, Apr 02 2013
(Sage)
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
(Magma) [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

Crossrefs

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

Sequence in context: A363838 A280117 A129364 * A035482 A322716 A372240

Adjacent sequences: A092284 A092285 A092286 * A092288 A092289 A092290

Keyword

nonn

Author

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

Extensions

Recurrence formula corrected by Daniel Forgues, Apr 07 2013

Status

approved