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
Examples
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
Programs
-
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] -
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
Comments