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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.
0
1, 24, 30, 90, 96, 342, 744, 812160, 330, 147258, 32784, 3314062080, 25896, 565632, 116412, 210317184, 145176, 6182491392, 963108
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
Similar sequence: A349478.
Sequence in context: A333122 A135045 A109321 * A348630 A067952 A239801
KEYWORD
nonn,more
AUTHOR
Amiram Eldar, Nov 25 2021
STATUS
approved