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A349503 a(n) is the least number k such that the continued fraction of the harmonic mean of the divisors of k contains n elements that are all distinct.

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%I A349503 #8 Nov 20 2021 20:49:57
%S A349503 1,2,20,52,156,768,8244,25808,406764,3610688,41395016,453695175,
%T A349503 3325792768
%N A349503 a(n) is the least number k such that the continued fraction of the harmonic mean of the divisors of k contains n elements that are all distinct.
%e A349503 The elements of the continued fractions of the harmonic mean of the divisors of the first 13 terms:
%e A349503    n        a(n)  elements
%e A349503   --  ----------  -----------------------------
%e A349503    1           1  1
%e A349503    2           2  1,3
%e A349503    3          20  2,1,6
%e A349503    4          52  3,5,2,4
%e A349503    5         156  4,1,3,2,5
%e A349503    6         768  6,1,3,4,2,13
%e A349503    7        8244  7,11,8,3,1,13,2
%e A349503    8       25808  5,6,3,13,1,2,4,7
%e A349503    9      406764  7,8,3,6,9,2,1,4,12
%e A349503   10     3610688  7,18,5,2,3,6,1,4,13,11
%e A349503   11    41395016  7,1,12,8,4,2,3,5,19,6,10
%e A349503   12   453695175  16,5,8,1,10,48,7,13,2,3,6,4
%e A349503   13  3325792768  19,1,21,7,6,3,12,13,5,9,2,8,4
%t A349503 cflen[n_] := Module[{cf = ContinuedFraction[DivisorSigma[0, n]/DivisorSigma[-1, n]], len}, If[(len = Length[cf]) == Length[DeleteDuplicates[cf]], len, 0]]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = cflen[n]; If[i > 0 && i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; TakeWhile[s, # > 0 &]]; seq[10, 10^7]
%Y A349503 Cf. A349473, A349502.
%K A349503 nonn,more
%O A349503 1,2
%A A349503 _Amiram Eldar_, Nov 20 2021