A349690 Numbers k such that the continued fraction of the abundancy index of k contains distinct elements.
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, 47, 49, 53, 56, 59, 60, 61, 67, 71, 73, 77, 79, 80, 81, 83, 88, 89, 91, 97, 101, 103, 104, 107, 109, 113, 120, 121, 125, 127, 131, 137, 139, 145, 149, 151, 155, 157, 163
Offset: 1
Keywords
Examples
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. 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.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
c[n_] := ContinuedFraction[DivisorSigma[1, n]/n]; q[n_] := Length[(cn = c[n])] == Length[DeleteDuplicates[cn]]; Select[Range[200], q]
-
PARI
isok(k) = my(v=contfrac(sigma(k)/k)); #v == #Set(v); \\ Michel Marcus, Nov 25 2021
Comments