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Numbers k such that the continued fraction of the abundancy index of k contains distinct elements.
2

%I #12 Nov 28 2021 02:55:56

%S 1,2,3,5,6,7,9,11,12,13,17,18,19,20,23,25,27,28,29,31,33,37,40,41,43,

%T 47,49,53,56,59,60,61,67,71,73,77,79,80,81,83,88,89,91,97,101,103,104,

%U 107,109,113,120,121,125,127,131,137,139,145,149,151,155,157,163

%N Numbers k such that the continued fraction of the abundancy index of k contains distinct elements.

%C All the primes (A000040) are terms of this sequence, since the continued fraction of the abundancy index of a prime p is {1, p}.

%C All the multiply-perfect numbers (A007691) are terms of this sequence, since the continued fraction of their abundancy index contains a single element.

%H Amiram Eldar, <a href="/A349690/b349690.txt">Table of n, a(n) for n = 1..10000</a>

%e 2 is a term since the abundancy index of 2 is 3/2 = 1 + 1/2 and the elements of the continued fraction, {1, 2}, are different.

%e 4 is not a term since the abundancy index of 4 is 7/4 = 1 + 1/(1 + 1/3) and the elements of the continued fraction, {1, 1, 3}, are not distinct.

%t c[n_] := ContinuedFraction[DivisorSigma[1, n]/n]; q[n_] := Length[(cn = c[n])] == Length[DeleteDuplicates[cn]]; Select[Range[200], q]

%o (PARI) isok(k) = my(v=contfrac(sigma(k)/k)); #v == #Set(v); \\ _Michel Marcus_, Nov 25 2021

%Y Cf. A007691, A349502, A349685, A349686.

%K nonn

%O 1,2

%A _Amiram Eldar_, Nov 25 2021