A349691 a(n) is the least number k such that the continued fraction of the abundancy index of k contains n elements that are all distinct, or -1 if no such k exists.

1, 2, 9, 176, 155, 2450, 21500, 118993, 767700, 12409639, 56024339, 857777653, 8648737607

Offset

1,2

Comments

a(14) > 4*10^10, if it exists.

Links

Table of n, a(n) for n=1..13.

Example

The elements of the continued fractions of the abundancy index of the first 13 terms are:
   n        a(n)  elements
  --  ----------  -----------------------------
   1           1  1
   2           2  1,2
   3           9  1,2,4
   4         176  2,8,1,4
   5         155  1,4,5,3,2
   6        2450  2,6,9,8,1,4
   7       21500  2,4,3,1,6,9,5
   8      118993  1,6,5,2,13,3,10,4
   9      767700  3,7,4,6,12,10,5,1,2
  10    12409639  1,10,12,6,3,2,4,14,5,7
  11    56024339  1,6,12,4,8,5,9,3,7,10,2
  12   857777653  1,14,3,5,12,4,6,2,7,9,10,8
  13  8648737607  1,12,6,13,2,4,10,7,11,3,9,8,5

Mathematica

cflen[n_] := Module[{cf = ContinuedFraction[DivisorSigma[1, n]/n], len}, If[(len = Length[cf]) == Length[DeleteDuplicates[cf]], len, 0]]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = cflen[n]; If[i > 0 && i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; TakeWhile[s, # > 0 &]]; seq[9, 10^6]

Prog

(PARI) isok(k, n) = my(v=contfrac(sigma(k)/k)); (#v == n) && (#Set(v) == n);
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Nov 25 2021

Crossrefs

Cf. A349503, A349685, A349690.

Sequence in context: A237201 A053294 A199695 * A078524 A041795 A123625

Adjacent sequences: A349688 A349689 A349690 * A349692 A349693 A349694

Keyword

nonn,more

Author

Amiram Eldar, Nov 25 2021

Status

approved