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 A129365 a(n)=A092287(n)/A129364(n). 4

%I

%S 1,1,1,1,1,2,2,2,6,48,48,48,48,1536,207360,207360,207360,1105920,

%T 1105920,17694720,30098718720,15410543984640,15410543984640,

%U 481579499520,60197437440000,123284351877120000,29958097506140160000

%N a(n)=A092287(n)/A129364(n).

%C Conjectures: A) a(n) is always an integer. B) If p is a prime then p|a(n) if and only if p <= n/3. 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). The precise decomposition of a(n) into primes would follow from the following two conjectures: C) For each positive integer n and prime p, ordp(a(np),p)= ordp(a(np+1),p)= ordp(a(np+2),p)= . . . = ordp(a(np+p-1),p). D) Let b(n)=A004125(n). Then ordp(a(np),p)= b(n)+ b(floor(n/p))+ b(floor(n/p^2))+ b(floor(n/p^3))+ . . .. This is reminiscent of de Polignac's formula (also due to Legendre) for the prime factorization of n! (see the link).

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/De_Polignac%27s_formula">De Polignac's formula</a>.

%F a(n)=(product{j=1..n}product{k=1..n} gcd(j,k))/(product{j=1..n}product{d|j} d^(j/d)). a(n)=(product{j=1..n}product{k=1..n}gcd(j,k))/(product{k=1..n}(floor(n/k)!)^k).

%Y Cf. A004125, A092287, A129364.

%K easy,nonn

%O 1,6

%A _Peter Bala_, Apr 13 2007

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Last modified July 16 23:49 EDT 2019. Contains 325092 sequences. (Running on oeis4.)