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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.
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%I #12 Aug 28 2023 08:20:46

%S 1,2,9,176,155,2450,21500,118993,767700,12409639,56024339,857777653,

%T 8648737607

%N 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 a(14) > 4*10^10, if it exists.

%e The elements of the continued fractions of the abundancy index of the first 13 terms are:

%e n a(n) elements

%e -- ---------- -----------------------------

%e 1 1 1

%e 2 2 1,2

%e 3 9 1,2,4

%e 4 176 2,8,1,4

%e 5 155 1,4,5,3,2

%e 6 2450 2,6,9,8,1,4

%e 7 21500 2,4,3,1,6,9,5

%e 8 118993 1,6,5,2,13,3,10,4

%e 9 767700 3,7,4,6,12,10,5,1,2

%e 10 12409639 1,10,12,6,3,2,4,14,5,7

%e 11 56024339 1,6,12,4,8,5,9,3,7,10,2

%e 12 857777653 1,14,3,5,12,4,6,2,7,9,10,8

%e 13 8648737607 1,12,6,13,2,4,10,7,11,3,9,8,5

%t 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 (PARI) isok(k, n) = my(v=contfrac(sigma(k)/k)); (#v == n) && (#Set(v) == n);

%o a(n) = my(k=1); while (!isok(k, n), k++); k; \\ _Michel Marcus_, Nov 25 2021

%Y Cf. A349503, A349685, A349690.

%K nonn,more

%O 1,2

%A _Amiram Eldar_, Nov 25 2021