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%I #9 Nov 21 2021 05:08:59
%S 1,1,1,1,1,2,1,2,2,2,1,1,1,2,2,1,1,1,1,1,1,2,1,3,1,2,1,3,1,3,1,3,1,1,
%T 1,1,1,1,1,1,1,2,1,3,2,1,1,1,1,2,1,2,1,1,3,1,1,1,1,2,1,1,1,1,2,1,1,3,
%U 1,1,1,2,1,1,1,1,1,1,1,3,1,1,1,2,1,1,1
%N a(n) is the smallest element in the continued fraction of the harmonic mean of the divisors of n.
%H Amiram Eldar, <a href="/A349497/b349497.txt">Table of n, a(n) for n = 1..10000</a>
%F a(p) = 1 for a prime p.
%F a(p^2) = 1 for a prime p != 3.
%F a(A129521(n)) = 1 for n > 3.
%F For a harmonic number m = A001599(k), a(m) = A099377(m) = A001600(k).
%e a(2) = 1 since the continued fraction of the harmonic mean of the divisors of 2, 4/3 = 1 + 1/3, has 2 elements, {1, 3}, and the smallest of them is 1.
%t a[n_] := Min[ContinuedFraction[DivisorSigma[0, n] / DivisorSigma[-1, n]]]; Array[a, 100]
%Y Cf. A001248, A001599, A001600, A099377, A099378, A129521, A349473.
%K nonn
%O 1,6
%A _Amiram Eldar_, Nov 20 2021
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