A153775
Sequence S such that 1 is in S and if x is in S, then 3x-1 and 3x are in S.
Original entry on oeis.org
1, 2, 3, 5, 6, 8, 9, 14, 15, 17, 18, 23, 24, 26, 27, 41, 42, 44, 45, 50, 51, 53, 54, 68, 69, 71, 72, 77, 78, 80, 81, 122, 123, 125, 126, 131, 132, 134, 135, 149, 150, 152, 153, 158, 159, 161, 162, 203, 204, 206, 207, 212, 213, 215, 216, 230, 231, 233, 234, 239, 240, 242
Offset: 1
See also the related sequences listed in
A191106.
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nxt[n_] := Flatten[3 # + {-1, 0} & /@ n]; Union[Flatten[NestList[nxt,{1},5]]] (* G. C. Greubel, Aug 28 2016 *)
A307744
A fractal function, related to ruler functions and the Cantor set. a(1) = 0; for m >= 0, a(3m) = 1; for m >= 1, a(3m-1) = a(m-1) + sign(a(m-1)), a(3m+1) = a(m+1) + sign(a(m+1)).
Original entry on oeis.org
1, 0, 2, 1, 3, 0, 1, 2, 3, 1, 4, 2, 1, 0, 4, 1, 2, 0, 1, 3, 2, 1, 4, 3, 1, 2, 4, 1, 5, 2, 1, 3, 5, 1, 2, 3, 1, 0, 2, 1, 5, 0, 1, 2, 5, 1, 3, 2, 1, 0, 3, 1, 2, 0, 1, 4, 2, 1, 3, 4, 1, 2, 3, 1, 5, 2, 1, 4, 5, 1, 2, 4, 1, 3, 2, 1, 5, 3, 1, 2, 5, 1, 6, 2, 1
Offset: 0
As 4 is congruent to 1 modulo 3, a(4) = a(3*1 + 1) = a(1+1) + sign (a(1+1)) = a(2) + sign(a(2)).
As 2 is congruent to -1 modulo 3, a(2) = a(3*1 - 1) = a(1-1) + sign (a(1-1)) = a(0) + sign(a(0)).
As 0 is congruent to 0 modulo 3, a(0) = 1. So a(2) = a(0) + sign(a(0)) = 1 + 1 = 2. So a(4) = a(2) + sign(a(2)) = 2 + 1 = 3.
For any m, the sequence from 9m - 9 to 9m + 9 can be represented by the table below. x, y and z represent distinct integers unless m = 0, in which case x = z = 0. Distinct values are shown in their own column to highlight patterns.
n a(n)
9m-9 1
9m-8 y - starts pattern (9m-8, 9m-4, 9m+4, 9m+8)
9m-7 2
9m-6 1
9m-5 x
9m-4 y
9m-3 1
9m-2 2
9m-1 x - ends pattern (9m-17, 9m-13, 9m-5, 9m-1)
9m 1
9m+1 z - starts pattern (9m+1, 9m+5, 9m+13, 9m+17)
9m+2 2
9m+3 1
9m+4 y
9m+5 z
9m+6 1
9m+7 2
9m+8 y - ends pattern (9m-8, 9m-4, 9m+4, 9m+8)
9m+9 1
For all m, one of x, y, z represents 3 in this table. Note the identical patterns indicated for "x", "y", "z" quadruples, and how the "x" quadruple ends 2 before the "z" quadruple starts, with the "y" quadruple overlapping both. For k >= 1, there are equivalent 2^k-tuples that overlap similarly, notably (3m-2, 3m+2) for all m.
Larger 2^k-tuples look more fractal, more obviously related to the Cantor set. See the pin plot of a(0..162) aligned above an inverted plot of ruler function A051064 in the links. 0's are emphasized with a fainter line running off the top of the plot, partly because 0 is used here as a conventional value and occurs with some properties (such as zero asymptotic density) that could be considered appropriate to the largest rather than smallest value in the sequence.
The table below illustrates the symmetries of scale of this sequence and ruler function A051064. Note the column for this sequence is indexed by k+1, 3k+1, 9k+1, whereas that for A051064 is indexed by k, 3k, 9k.
a(n+1) A051064(n)
n=k, k=16..27 0,1,3,2,1,4,3,1,2,4,1,5 1,1,3,1,1,2,1,1,2,1,1,4
n=3k,k=16..27 0,2,4,3,2,5,4,2,3,5,2,6 2,2,4,2,2,3,2,2,3,2,2,5
n=9k,k=16..27 0,3,5,4,3,6,5,3,4,6,3,7 3,3,5,3,3,4,3,3,4,3,3,6
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a(n) = if (n==1, 0, my(m=n%3); if (m==0, 1, my(kk = (if (m==1, a(n\3+1), a((n-2)\3)))); kk + sign(kk)));
for (n=0, 100, print1(a(n), ", ")) \\ Michel Marcus, Jul 06 2019
A308364
a(0) = 0, a(3n) = a(n), a(3n+1) = a(n)*3 + 1, a(3n-1) = a(n)*3 - 1.
Original entry on oeis.org
0, 1, 2, 1, 4, 5, 2, 7, 2, 1, 4, 11, 4, 13, 14, 5, 16, 5, 2, 7, 20, 7, 22, 5, 2, 7, 2, 1, 4, 11, 4, 13, 32, 11, 34, 11, 4, 13, 38, 13, 40, 41, 14, 43, 14, 5, 16, 47, 16, 49, 14, 5, 16, 5, 2, 7, 20, 7, 22, 59, 20, 61, 20, 7, 22, 65, 22, 67, 14, 5, 16, 5, 2, 7, 20, 7, 22, 5, 2, 7, 2, 1
Offset: 0
As 6 is congruent to 0 modulo 3, a(6) = a(3*2) = a(2).
As 2 is congruent to -1 modulo 3, a(2) = a(3*1 - 1) = a(1)*3 - 1.
As 1 is congruent to 1 modulo 3, a(1) = a(0*1 + 1) = a(0)*3 + 1 = 0*3 + 1 = 1.
So a(2) = a(1)*3 - 1 = 1*3 - 1 = 2. So a(6) = a(2) = 2.
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a:=[1]; for n in [2..80] do if n mod 3 eq 2 then a[n]:= 3*a[(n+1) div 3]-1; end if; if n mod 3 eq 1 then a[n]:=3*a[(n-1) div 3]+1; end if; if n mod 3 eq 0 then a[n]:=a[n div 3]; end if; end for; [0] cat a; // Marius A. Burtea, Nov 14 2019
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a(n) = if (n == 0, 0, r = n%3; if (r==0, a(n/3), if (r==1, 3*a((n-1)/3)+1, 3*a((n+1)/3)-1))); \\ Michel Marcus, May 29 2019
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