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A110566 a(n) = lcm{1,2,...,n}/denominator of harmonic number H(n).

%I A110566 #39 May 17 2023 10:29:56
%S A110566 1,1,1,1,1,3,3,3,1,1,1,1,1,1,1,1,1,3,3,15,45,45,45,15,3,3,1,1,1,1,1,1,
%T A110566 11,11,11,11,11,11,11,11,11,77,77,7,7,7,7,7,1,1,1,1,1,3,3,3,3,3,3,3,3,
%U A110566 3,9,9,9,27,27,27,9,9,9,3,3,3,3,3,33,33,33,33,11,11,11,11,11,11,11,1,1,1
%N A110566 a(n) = lcm{1,2,...,n}/denominator of harmonic number H(n).
%C A110566 a(n) is always odd.
%C A110566 Unsorted union: 1, 3, 15, 45, 11, 77, 7, 9, 27, 33, 25, 5, 55, 275, 13, 39, 17, 49, 931, 19, 319, 75, ..., . See A112810.
%C A110566 It is conjectured that every odd number occurs in this sequence (see A112822 for the first occurrence of each of them). - _Jianing Song_, Nov 28 2022
%H A110566 Alois P. Heinz, <a href="/A110566/b110566.txt">Table of n, a(n) for n = 1..10000</a>
%F A110566 a(n) = A003418(n)/A002805(n) = A025529(n)/A001008(n).
%F A110566 From _Franz Vrabec_, Sep 21 2005: (Start)
%F A110566 a(n) = gcd(lcm{1,2,...,n}, H(n)*lcm{1,2,...,n}).
%F A110566 a(n) = gcd(A003418(n), A025529(n)). (End)
%e A110566 a(6) = 60/20 = 3 because lcm{1,2,3,4,5,6}=60 and H(6)=49/20.
%p A110566 H:= proc(n) H(n):= 1/n +`if`(n=1, 0, H(n-1)) end:
%p A110566 L:= proc(n) L(n):= ilcm(n, `if`(n=1, 1, L(n-1))) end:
%p A110566 a:= n-> L(n)/denom(H(n)):
%p A110566 seq(a(n), n=1..100);  # _Alois P. Heinz_, Aug 30 2012
%t A110566 f[n_] := LCM @@ Range[n]/Denominator[HarmonicNumber[n]]; Table[ f[n], {n, 90}] (* _Robert G. Wilson v_, Sep 15 2005 *)
%o A110566 (PARI) a(n) = lcm(vector(n, k, k))/denominator(sum(k=1, n, 1/k)); \\ _Michel Marcus_, Mar 07 2018
%o A110566 (Python)
%o A110566 from sympy import lcm, harmonic
%o A110566 def A110566(n): return lcm([k for k in range(1,n+1)])//harmonic(n).q # _Chai Wah Wu_, Mar 06 2021
%Y A110566 Cf. A001008, A002805, A003418, A025529, A098464, A112810, A112822, A358557.
%K A110566 nonn
%O A110566 1,6
%A A110566 _Franz Vrabec_, Sep 12 2005
%E A110566 More terms from _Robert G. Wilson v_, Sep 15 2005