A129454 a(n) = Product{i=1..n-1} Product{j=1..n-1} Product{k=1..n-1} gcd(i,j,k).

1, 1, 1, 2, 6, 1536, 7680, 8806025134080, 61642175938560, 2168841254587541957294161920, 7562281854741110985626291951024209920, 1362299589723309231779453337910253309054734620740812800000000

Offset: 0

Author

Peter Bala, Apr 16 2007

Keywords

• nonn

Comments

Conjecture: Let p be a prime and let ordp(n,p) denote the largest power of p which 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 ordp(a(n),p)=(floor(n/p))^3 + (floor(n/p^2))^3 + (floor(n/p^3))^3 + . . .. Compare with the comments in A092287.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..26

Crossrefs

Cf. A092287, A129455.

Programs

Magma

A129454:= func< n | n le 1 select 1 else (&*[(&*[(&*[GCD(GCD(j,k),m): k in [1..n-1]]): j in [1..n-1]]): m in [1..n-1]]) >;
[A129454(n): n in [0..20]]; // G. C. Greubel, Feb 07 2024

Mathematica

A129454[n_]:= Product[GCD[j,k,m], {j,n-1}, {k,n-1}, {m,n-1}];
Table[A129454[n], {n,0,20}] (* G. C. Greubel, Feb 07 2024 *)

SageMath

def A129454(n): return product(product(product(gcd(gcd(j,k),m) for k in range(1,n)) for j in range(1,n)) for m in range(1,n))
[A129454(n) for n in range(21)] # G. C. Greubel, Feb 07 2024

Formula

a(n) = Product{i=1..n-1} Product{j=1..n-1} Product{k=1..n-1} gcd(i,j,k), for n > 2, otherwise a(n) = 1.