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%I #66 Feb 07 2024 02:53:44
%S 1,1,2,6,96,480,414720,2903040,5945425920,4334215495680,
%T 277389791723520000,3051287708958720000,437332621360674939863040000,
%U 5685324077688774218219520000,15974941971638268369709427589120000,982608696336737613503095822614528000000000
%N a(n) = Product_{j=1..n} Product_{k=1..n} gcd(j,k).
%C 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
%C The conjecture is correct. - _Charles R Greathouse IV_, Apr 02 2013
%C 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
%C 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
%H T. D. Noe, <a href="/A092287/b092287.txt">Table of n, a(n) for n = 0..67</a>
%F Also a(n) = Product_{k=1..n} Product_{j=1..n} lcm(1..floor(min(n/k, n/j))).
%F From _Daniel Forgues_, Apr 08 2013: (Start)
%F 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).
%F Formula for n >= 0: a(n) = n! * (Product_{j=1..n} Product_{k=1..j-1} gcd(j,k))^2. (End)
%F a(n) = n! * A224479(n)^2 (the last formula above).
%F a(n) = n$ * A224497(n)^4, n$ the swinging factorial A056040(n). - _Peter Luschny_, Apr 10 2013
%p f := n->mul(mul(igcd(j,k),k=1..n),j=1..n);
%t 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_ *)
%o (PARI) h(n,p)=if(n<p,0,n\=p;h(n,p)+n^2)
%o a(n)=prod(i=1,primepi(n),my(p=prime(i));p^h(n,p)) \\ _Charles R Greathouse IV_, Apr 02 2013
%o (Sage)
%o def A092287(n):
%o R = 1
%o for p in primes(n+1) :
%o s = 0; r = n
%o while r > 0 :
%o r = r//p
%o s += r*r
%o R *= p^s
%o return R
%o [A092287(i) for i in (0..15)] # _Peter Luschny_, Apr 10 2013
%o (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
%Y Cf. A003989, A018806, A051190, A090494, A129365, A129439, A129453, A129454, A129455, A224479, A224497.
%K nonn
%O 0,3
%A _N. J. A. Sloane_, based on a suggestion from _Leroy Quet_, Feb 03 2004
%E Recurrence formula corrected by _Daniel Forgues_, Apr 07 2013
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