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%I #8 Feb 07 2024 09:01:57
%S 1,1,1,2,6,1536,7680,8806025134080,61642175938560,
%T 2168841254587541957294161920,7562281854741110985626291951024209920,
%U 1362299589723309231779453337910253309054734620740812800000000
%N a(n) = Product{i=1..n-1} Product{j=1..n-1} Product{k=1..n-1} gcd(i,j,k).
%C 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.
%H G. C. Greubel, <a href="/A129454/b129454.txt">Table of n, a(n) for n = 0..26</a>
%F 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.
%t A129454[n_]:= Product[GCD[j,k,m], {j,n-1}, {k,n-1}, {m,n-1}];
%t Table[A129454[n], {n,0,20}] (* _G. C. Greubel_, Feb 07 2024 *)
%o (Magma)
%o 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]]) >;
%o [A129454(n): n in [0..20]]; // _G. C. Greubel_, Feb 07 2024
%o (SageMath)
%o 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))
%o [A129454(n) for n in range(21)] # _G. C. Greubel_, Feb 07 2024
%Y Cf. A092287, A129455.
%K nonn
%O 0,4
%A _Peter Bala_, Apr 16 2007
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