A121153
Numbers k with the property that 1/k can be written in base 3 in such a way that the fractional part contains no 1's.
Original entry on oeis.org
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
Offset: 1
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.
-
(* 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] &]
-
is(n,R=divrem(3^logint(n,3),n),S=0)={while(R[1]!=1&&!bittest(S,R[2]), S+=1<M. F. Hasler, Feb 27 2018
A170943
Numbers n with the property that when 1/n is written in base 3 (in either of the two representations, if the representation is ambiguous) the fractional part contains no 1's.
Original entry on oeis.org
1, 4, 10, 12, 13, 28, 30, 36, 39, 40, 82, 84, 90, 91, 108, 117, 120, 121, 244, 246, 252, 270, 273, 324, 328, 351, 360, 363, 364, 730, 732, 738, 756, 757, 810, 819, 820, 949, 972, 984, 1036, 1053, 1080, 1089, 1092, 1093, 2188, 2190, 2196, 2214, 2268, 2271, 2362, 2430
Offset: 1
1/3 in base 3 can be written as either .1 or .0222222... The first version contains a 1, so 3 is not in the sequence.
1/4 in base 3 is .02020202020..., so 4 is in the sequence.
Original entry on oeis.org
2, 5, 6, 7, 8, 11, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 32, 33, 34, 35, 37, 38, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 83, 85, 86, 87, 88, 89
Offset: 1
A170952
Take the Cantor set sequence A121153 and if the entry m = A121153(n) is in the range 3^k <= m < 3^(k+1), subtract 3^k from it.
Original entry on oeis.org
0, 0, 1, 0, 1, 3, 4, 0, 1, 3, 9, 12, 13, 0, 1, 3, 9, 10, 27, 36, 39, 40, 0, 1, 3, 9, 27, 30, 81, 85, 108, 117, 120, 121, 0, 1, 3, 9, 27, 28, 81, 90, 91, 220, 243, 255, 307, 324, 351, 360, 363, 364, 0, 1, 3, 9, 27, 81, 84, 175, 243, 270, 273, 625, 660, 729, 733, 765, 921, 972, 1053
Offset: 1
If written as a triangle:
0,
0, 1,
0, 1, 3, 4,
0, 1, 3, 9, 12, 13,
0, 1, 3, 9, 10, 27, 36, 39, 40,
0, 1, 3, 9, 27, 30, 81, 85, 108, 117, 120, 121,
0, 1, 3, 9, 27, 28, 81, 90, 91, 220, 243, 255, 307, 324, 351, 360, 363, 364,
0, 1, 3, 9, 27, 81, 84, 175, 243, 270, 273, 625, 660, 729, 733, 765, 921, 972, 1053, 1080, 1089, 1092, 1093,
...
A368558
Number of fractions i/n that are in the Cantor set.
Original entry on oeis.org
2, 2, 4, 4, 2, 4, 2, 4, 8, 6, 2, 8, 8, 2, 4, 4, 2, 8, 2, 8, 4, 2, 2, 8, 2, 8, 16, 10, 2, 12, 2, 4, 4, 2, 2, 16, 2, 2, 16, 16, 2, 4, 2, 4, 8, 2, 2, 8, 2, 6, 4, 10, 2, 16, 2, 10, 4, 2, 2, 16, 2, 2, 8, 4, 8, 4, 2, 4, 4, 6, 2, 16, 2, 2, 4, 4, 2, 16, 2, 16
Offset: 1
For n = 12, there are a(12) = 8 fractions, and their numerators are i = 0, 1, 3, 4, 8, 9, 11, 12.
- Jason Yuen, Table of n, a(n) for n = 1..10000
- 2023 Canadian Computing Competition Senior Committee, 2023 CCC Senior Problems, Problem S5: The Filter, University of Waterloo.
- 2023 Canadian Computing Competition Senior Committee, 2023 CCC Senior Problem Commentary, S5 The Filter, University of Waterloo.
- Zixiang Zhou, main.cpp. This C++ program solves CCC 2023, Problem S5 with a time complexity of O(n^(log_3(2)) log n).
-
def is_member(i, n): # Returns True if i/n is in the Cantor set
visited = set()
while True:
if n < 3 * i < 2 * n: return False
if i in visited: return True
visited.add(i)
i = 3 * min(i, n - i)
def a(n): return sum(is_member(i, n) for i in range(n + 1))
Showing 1-5 of 5 results.
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