OFFSET
1,2
COMMENTS
a(21) = 7094832, a(23) = 24167070, a(25) = 858983598, a(27) = 1137635260, a(29) = 1402857468, a(31) = 45230309244, and there are no more terms below 1.6*10^11.
EXAMPLE
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
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]
PROG
(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
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Amiram Eldar, Nov 25 2021
STATUS
approved