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Denominator(Sum_{i=1..n} 1/i^3)/denominator(Sum_{i=1..n} 1/i).
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%I #15 May 07 2020 06:16:06

%S 1,4,36,144,3600,1200,58800,235200,6350400,6350400,768398400,

%T 768398400,129859329600,129859329600,129859329600,519437318400,

%U 150117385017600,50039128339200,18064125330451200,18064125330451200,54192375991353600

%N Denominator(Sum_{i=1..n} 1/i^3)/denominator(Sum_{i=1..n} 1/i).

%C For n = 1 to n = 19, we have a(n) = A334580(n), but a(20) = 18064125330451200 <> 3612825066090240 = A334580(20). - _Petros Hadjicostas_, May 06 2020

%F a(n) = A007409(n)/A002805(n).

%p a := proc(n) local i: denom(add(1/i^3, i = 1 .. n))/denom(add(1/i, i = 1 .. n)): end proc:

%p seq(a(n), n=1..30); # _Petros Hadjicostas_, May 06 2020

%o (PARI) a(n) = denominator(sum(k=1, n, 1/k^3)/sum(k=1, n, 1/k)); \\ _Michel Marcus_, May 07 2020

%Y Cf. A002805, A007409, A119682, A334580.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, Mar 27 2002