A349689 a(n) is the least number k such that the sequence of elements of the abundancy index of k is palindromic with length n, or -1 if no such k exists.
1, 24, 30, 90, 96, 342, 744, 812160, 330, 147258, 32784, 3314062080, 25896, 565632, 116412, 210317184, 145176, 6182491392, 963108
Offset: 1
Examples
The elements of the continued fractions of the abundancy index of the terms are: n a(n) elements -- -------- ------------------------------------- 1 1 1 2 24 2,2 3 30 2,2,2 4 90 2,1,1,2 5 96 2,1,1,1,2 6 342 2,3,1,1,3,2 7 744 2,1,1,2,1,1,2 8 812160 3,1,1,1,1,1,1,3 9 330 2,1,1,1,1,1,1,1,2 10 147258 2,3,1,2,5,5,2,1,3,2 11 32784 2,1,1,2,2,1,2,2,1,1,2 12 3314062080 4,2,1,1,2,1,1,2,1,1,2,4 13 25896 2,1,2,1,1,1,2,1,1,1,2,1,2 14 565632 2,1,7,1,1,2,1,1,2,1,1,7,1,2 15 116412 2,2,1,1,1,1,1,8,1,1,1,1,1,2,2 16 210317184 3,1,1,2,3,3,2,1,1,2,3,3,2,1,1,3 17 145176 2,1,1,1,1,1,1,1,4,1,1,1,1,1,1,1,2 18 6182491392 3,1,1,2,7,3,2,2,1,1,2,2,3,7,2,1,1,3 19 963108 2,1,1,1,1,2,1,1,1,3,1,1,1,2,1,1,1,1,2
Programs
-
Mathematica
cfai[n_] := ContinuedFraction[DivisorSigma[1, n]/n]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i, cf}, While[c < len && n < nmax, cf = cfai[n]; If[PalindromeQ[cf] && (i = Length[cf]) <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; TakeWhile[s, # > 0 &]]; seq[11, 10^6] -
PARI
isok(k, n) = my(v=contfrac(sigma(k)/k)); (#v == n) && (v == Vecrev(v)); a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Nov 25 2021
Comments