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A349686 Numbers k such that the continued fraction of the abundancy index of k contains a single distinct element. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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?

LINKS

Table of n, a(n) for n=1..38.

EXAMPLE

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}.

MATHEMATICA

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

PROG

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

CROSSREFS

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

Sequence in context: A072710 A349688 A273124 * A069235 A175200 A293453

Adjacent sequences:  A349683 A349684 A349685 * A349687 A349688 A349689

KEYWORD

nonn

AUTHOR

Amiram Eldar, Nov 25 2021

STATUS

approved

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Last modified August 13 07:13 EDT 2022. Contains 356078 sequences. (Running on oeis4.)