A349686 -id:A349686 - cp's OEIS frontend

A349688 Numbers k such that the sequence of elements of the continued fraction of the abundancy index of k is palindromic.

1, 6, 24, 28, 30, 42, 54, 66, 70, 78, 84, 90, 96, 102, 114, 120, 138, 140, 174, 186, 220, 222, 246, 258, 264, 270, 282, 308, 318, 330, 342, 348, 354, 364, 366, 402, 426, 438, 474, 476, 496, 498, 532, 534, 582, 606, 618, 642, 644, 654, 660, 672, 678, 744, 760, 762

Offset: 1

Amiram Eldar, Nov 25 2021

nonn

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

			24 is a term since the sequence of elements of the abundancy index of 24, sigma(24)/24 = 5/2 = 2 + 1/2, is {2, 2}, which is palindromic.
42 is a term since the sequence of elements of the abundancy index of 42, sigma(42)/42 = 16/7 = 2 + 1/(3 + 1/2), is {2, 3, 2}, which is palindromic.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A349685.
A007691 and A349686 are subsequences.
Similar sequence: A349477.

q[n_] := PalindromeQ[ContinuedFraction[DivisorSigma[1, n]/n]]; Select[Range[1000], q]

isok(k) = my(v=contfrac(sigma(k)/k)); v == Vecrev(v); \\ Michel Marcus, Nov 25 2021

A349690 Numbers k such that the continued fraction of the abundancy index of k contains distinct elements.

Original entry on oeis.org

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

Amiram Eldar, Nov 25 2021

nonn

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

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

			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.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A007691, A349502, A349685, A349686.

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

isok(k) = my(v=contfrac(sigma(k)/k)); #v == #Set(v); \\ Michel Marcus, Nov 25 2021

cp's OEIS Frontend

A349688 Numbers k such that the sequence of elements of the continued fraction of the abundancy index of k is palindromic.

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