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 A121153 Numbers n with the property that 1/n can be written in base 3 in such a way that the fractional part contains no 1's. 12
 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 36, 39, 40, 81, 82, 84, 90, 91, 108, 117, 120, 121, 243, 244, 246, 252, 270, 273, 324, 328, 351, 360, 363, 364, 729, 730, 732, 738, 756, 757, 810, 819, 820, 949, 972, 984, 1036, 1053, 1080, 1089, 1092, 1093, 2187 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Numbers n such that 1/n is in the Cantor set. A subsequence of A054591. The first member of A054591 which does not belong to this sequence is 146. See A135666. This is not a subsequence of A005836 (949 belongs to the present sequence but not to A005836). See A170830, A170853. LINKS T. D. Noe and N. J. A. Sloane, Table of n, a(n) for n=1..1164 (terms < 3^21) D. Jordan and R. Schayer Rational points on the Cantor middle thirds set [Broken link corrected by Rainer Rosenthal, Feb 20 2009] EXAMPLE 1/3 in base 3 can be written as either .1 or .0222222... The latter version contains no 1's, so 3 is in the sequence. 1/4 in base 3 is .02020202020..., so 4 is in the sequence. MATHEMATICA (Mma code from T. D. Noe, Feb 20 2010. This produces the sequence except for the powers of 3.) # Find the length of the periodic part of the fraction: FracLen[n_] := Module[{r = n/3^IntegerExponent[n, 3]}, MultiplicativeOrder[3, r]] # Generate the fractions and select those that have no 1's: Select[Range[100000], ! MemberQ[Union[RealDigits[1/#, 3, FracLen[ # ]][[1]]], 1] &] PROG (PARI) is(n, R=divrem(3^logint(n, 3), n), S=0)={while(R[1]!=1&&!bittest(S, R[2]), S+=1<

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Last modified June 16 21:15 EDT 2021. Contains 345080 sequences. (Running on oeis4.)