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A349691 a(n) is the least number k such that the continued fraction of the abundancy index of k contains n elements that are all distinct, or -1 if no such k exists.

%I A349691 #12 Aug 28 2023 08:20:46
%S A349691 1,2,9,176,155,2450,21500,118993,767700,12409639,56024339,857777653,
%T A349691 8648737607
%N A349691 a(n) is the least number k such that the continued fraction of the abundancy index of k contains n elements that are all distinct, or -1 if no such k exists.
%C A349691 a(14) > 4*10^10, if it exists.
%e A349691 The elements of the continued fractions of the abundancy index of the first 13 terms are:
%e A349691    n        a(n)  elements
%e A349691   --  ----------  -----------------------------
%e A349691    1           1  1
%e A349691    2           2  1,2
%e A349691    3           9  1,2,4
%e A349691    4         176  2,8,1,4
%e A349691    5         155  1,4,5,3,2
%e A349691    6        2450  2,6,9,8,1,4
%e A349691    7       21500  2,4,3,1,6,9,5
%e A349691    8      118993  1,6,5,2,13,3,10,4
%e A349691    9      767700  3,7,4,6,12,10,5,1,2
%e A349691   10    12409639  1,10,12,6,3,2,4,14,5,7
%e A349691   11    56024339  1,6,12,4,8,5,9,3,7,10,2
%e A349691   12   857777653  1,14,3,5,12,4,6,2,7,9,10,8
%e A349691   13  8648737607  1,12,6,13,2,4,10,7,11,3,9,8,5
%t A349691 cflen[n_] := Module[{cf = ContinuedFraction[DivisorSigma[1, n]/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[9, 10^6]
%o A349691 (PARI) isok(k, n) = my(v=contfrac(sigma(k)/k)); (#v == n) && (#Set(v) == n);
%o A349691 a(n) = my(k=1); while (!isok(k, n), k++); k; \\ _Michel Marcus_, Nov 25 2021
%Y A349691 Cf. A349503, A349685, A349690.
%K A349691 nonn,more
%O A349691 1,2
%A A349691 _Amiram Eldar_, Nov 25 2021