A349686 Numbers k such that the continued fraction of the abundancy index of k contains a single distinct element.

1, 6, 24, 28, 30, 120, 140, 348, 496, 672, 1080, 2480, 6048, 6200, 6552, 6786, 8128, 30240, 32760, 40640, 143880, 238080, 435708, 514080, 523776, 524160, 556920, 805728, 1997868, 2178540, 4713984, 23569920, 33550336, 37035180, 38958426, 45532800, 91963648, 142990848

Offset: 1

Author

Amiram Eldar, Nov 25 2021

Keywords

• nonn

Comments

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

Up to 4*10^10 the continued fractions of the abundancy indices of the terms have lengths 1, 2, 3, 5 or 11. The least terms that are corresponding to these lengths are 1, 24, 30, 348 and 1997868, respectively. Are there terms with other lengths?

Examples

			24 is a term since the continued fraction of its abundancy index sigma(24)/24 = 5/2 = 2 + 1/2 has the elements {2, 2}.
30 is a term since the continued fraction of its abundancy index sigma(30)/30 = 12/5 = 2 + 1/(2 + 1/2) has the elements {2, 2, 2}.
143880 is a term since the continued fraction of its abundancy index sigma(143880)/143880 = 360/109 = 3 + 1/(3 + 1/(3 + 1/(3 + 1/3))) has the elements {3, 3, 3}.
		

Crossrefs

Cf. A000203, A007691, A071862, A071865, A017665, A017666, A349685.

Programs

Mathematica

c[n_] := ContinuedFraction[DivisorSigma[1, n] / n]; q[n_] := Length[Union[c[n]]] == 1; Select[Range[10^6], q]

PARI

isok(k) = #Set(contfrac(sigma(k)/k)) == 1; \\ Michel Marcus, Nov 25 2021