cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A087088 Positive ruler-type fractal sequence with 1's in every third position.

Original entry on oeis.org

1, 2, 3, 1, 4, 2, 1, 5, 3, 1, 2, 6, 1, 4, 2, 1, 3, 7, 1, 2, 5, 1, 3, 2, 1, 4, 8, 1, 2, 3, 1, 6, 2, 1, 4, 3, 1, 2, 5, 1, 9, 2, 1, 3, 4, 1, 2, 7, 1, 3, 2, 1, 5, 4, 1, 2, 3, 1, 6, 2, 1, 10, 3, 1, 2, 4, 1, 5, 2, 1, 3, 8, 1, 2, 4, 1, 3, 2, 1, 6, 5, 1, 2, 3, 1, 4, 2, 1, 7, 3, 1, 2, 11, 1, 4, 2, 1, 3, 5, 1, 2, 6, 1, 3, 2
Offset: 1

Views

Author

Enrico T. Federighi (rico125162(AT)aol.com), Aug 08 2003

Keywords

Comments

If all the terms in the sequence are reduced by one and then all zeros are removed, the result is the same as the original sequence.
From Benoit Cloitre, Mar 07 2009: (Start)
To construct the sequence:
Step 1: start from a sequence of 1's and leave two undefined places between every pair of 1's giving: 1,(),(),1,(),(),1,(),(),1,(),(),1,(),(),1,...
Step 2: replace the first undefined place with a 2 and henceforth leave two undefined places between two 2's giving: 1,2,(),1,(),2,1,(),(),1,2,(),1,(),2,1,...
Step 3: replace the first undefined place with a 3 and henceforth leave two undefined places between two 3's giving: 1,2,3,1,(),2,1,(),3,1,2,(),1,(),2,1,...
Step 4: replace the first undefined place with a 4 and leave 2 undefined places between two 4's giving: 1,2,3,1,4,2,1,(),3,1,2,(),1,4,2,1,... Iterating the process indefinitely yields the sequence: 1,2,3,1,4,2,1,5,3,1,2,6,1,4,2,1,... (End)
From Peter Munn, Jul 10 2020: (Start)
For k >= 1, the number k occurs in a pattern with fundamental period 3^k, and with points of mirror symmetry at intervals of (3^k)/2. Those points have an extrapolated common origin (for k >= 1) at an offset 1.5 to the left of the sequence's initial "1". The snake format illustration in the example section may be useful for visualizing this.
(End)
For k >= 1, k first occurs at position A061419(k) and its k-th occurrence is at position A083045(k-1). - Peter Munn, Aug 23 2020
(a(n)) is the unique fixed point of the two-block substitution a,b -> 1,a+1,b+1, where a,b are natural numbers. - Michel Dekking, Sep 26 2022

Examples

			From _Peter Munn_, Jul 03 2020: (Start)
Listing the terms in a snake format (with period 27) illustrates periodic and mirror symmetries. Horizontal lines mark points of mirror symmetry for 3's. Vertical lines mark further points of mirror symmetry for 2's. 79 terms are shown. (Referred to the extrapolated common origin of periodic mirror symmetry, the initial term is at offset 1.5 and the last shown is at offset 79.5 = 3^4 - 1.5.) Observe also mirror symmetry of 4's (seen vertically).
    1  2  3  1  4  2  1  5   3  1  2  6
             |             |            1 --
    1  2  3  1  5  2  1  7   3  1  2  4
_ 4
  8
    1  2  3  1  6  2  1  4   3  1  2  5
             |             |            1 --
    1  2  3  1  7  2  1  4   3  1  2  9
_ 5
  4
    1  2  3  1  6  2  1 10   3  1  2  4
             |             |            1 --
    1  2  3  1  4  2  1  8   3  1  2  5
(End)
From _Peter Munn_, Aug 22 2020: (Start)
The start of the sequence is shown below in conjunction with related sequences, aligning their points of mirror symmetry. The longer, and shorter, vertical lines indicate points of mirror symmetry for terms valued less than 4, and less than 3, respectively. Note each term of A051064 is the minimum of two terms displayed nearest below it, and each term of A254046 is the minimum of the two terms displayed diagonally above it.
        |                          |                          |
A051064:| 1 1 2 1 1 2 1 1 3 1 1 2 1 1 2 1 1 3 1 1 2 1 1 2 1 1 4 1 1 2
        |        |        |        |        |        |        |
[a(n)]: |  1 2 3 1 4 2 1 5 3 1 2 6 1 4 2 1 3 7 1 2 5 1 3 2 1 4 8 1 2 3
        |        |        |        |        |        |        |
A254046:|1 2 1 1 3 1 1 2 1 1 2 1 1 4 1 1 2 1 1 2 1 1 3 1 1 2 1 1 2 1 1
        |                          |                          |
(End)
		

Crossrefs

Sequences with equivalent symmetries: A051064, A254046.
Records are given by A061419: a(A061419(n))=n.
Essentially the odd bisection of A335933.
Sequence with similar definition: A087165.
Ordinal transform of A163491, with which this sequence has a joint relationship to A083044, A083045.
See also the comment in A024629.

Programs

Formula

a(n) = 1 when n == 1 (mod 3), otherwise a(n) = a(n-ceiling(n/3)) + 1.
a(n) = 3 + A244040(3*(n-1)) - A244040(3*n). - Tom Edgar and James Van Alstine, Aug 04 2014
From Peter Munn, Aug 22 2020: (Start)
For m >= 0, a(3*m+1) = 1; a(3*m+2) = a(2*m+1) + 1; a(3*m+3) = a(2*m+2) + 1.
For n >= 1, the following identities hold.
a(n) = A335933(2*n+1).
A083044(A163491(n) - 1, a(n) - 1) = n.
A051064(n+1) = min(a(n), a(n+1)).
A254046(n+2) = min(a(n), a(n+2)). (End)

Extensions

More terms from Paul D. Hanna, Aug 21 2003
Offset changed by M. F. Hasler (following remarks by Peter Munn), Jul 13 2020
Thanks to Allan C. Wechsler for suggesting the new name. - N. J. A. Sloane, Jul 14 2020

A205592 a(2) = 1, a(3k) = a(3k+1) = a(2k), a(3k+2) = 2a(2k+1) for k >= 1.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 4, 1, 1, 2, 4, 4, 2, 1, 1, 4, 4, 4, 8, 2, 2, 2, 1, 1, 8, 4, 4, 8, 8, 8, 4, 2, 2, 4, 1, 1, 2, 8, 8, 8, 4, 4, 16, 8, 8, 16, 4, 4, 4, 2, 2, 8, 1, 1, 2, 2, 2, 16, 8, 8, 16, 4, 4, 8, 16, 16, 16, 8, 8, 32, 4, 4, 8, 4, 4, 4, 2, 2, 16, 1, 1, 2, 2, 2, 4
Offset: 2

Views

Author

Joseph Myers, Jan 29 2012

Keywords

Crossrefs

Formula

a(n) = A205591(n) - A205591(n-1).

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

Views

Author

Peter Munn, Apr 26 2019

Keywords

Comments

The sequence extends to negative n by defining a(n) = a(-n).
For k >= 1 numbers 1..k occur with the same periodic and mirror symmetries as in ruler function A051064, in which k occurs 3 times more frequently than k+1. Here k occurs 3/2 times more frequently than k+1, precisely 2^(k-1) times in every 3^k terms. 0 has asymptotic density 0. Taking a trisection shows some scale symmetry, again comparable to ruler functions, as illustrated in the example section.
The links include a pin plot of a(0..162) aligned above an inverted plot of A051064 (the emphatic marking of 0's is significant). Between each n_k where A051064(n_k) = k >= 2 and the nearest n_k' where A051064(n_k') > k (or n_k' = 0 if nearer), there are 2^(k-2) indices where k occurs in this sequence, forming a 2^(k-2)-tuple. The 2^(k-2)-tuples have identical patterns and each has symmetry about an n_(k-1) where A051064(n_(k-1)) = k-1.
For a given k, the tuples described above are periodic with two per fundamental period, and the closest pairs of these tuples jointly form the pattern of one of the equivalent tuples for k+1. These patterns relate to the nonperiodic pattern for 0's and to the Cantor set as follows.
Let S_k be the sequence of positive indices at which k occurs, with 3^(k-2) subtracted when k >= 2. Given its ruler-type symmetries, S_k k >= 2 is determined by its first 2^(k-2) terms, which are the same as the first 2^(k-2) terms of S_i for i > k. The limiting sequence as k goes to infinity is S_0, which is A191108. {A191108(i)/(2*3^k) | 1 <= i <= 2^k} is the set of center points of the intervals deleted at step k+1 when generating the Cantor ternary set. This leads to the following scaling property.
Define c: Z -> P(R) so that c(n) is the scaled and translated Cantor ternary set spanning [n-1, n+1], and let C_k be the union of c(n) for all integer n with a(n) = k. Clearly C_1 consists of a scaled Cantor set repeated with period 3. (The set's two nonempty thirds occur at alternating intervals of 4/3 and 5/3.) For k >= 1, C_k is C_1 scaled by 3^(k-1), consisting therefore of a scaled Cantor set repeated with period 3^k. C_0 is special: C_0 = (C_0)*3 = (C_0)/3 = -C_0. Specifically, (C_0)/2 is the closure of the Cantor ternary set under multiplication by 3 and by -1.
Take a Sierpinski arrowhead curve formed of unit edges numbered consecutively from 0 at its axis of symmetry and aligned with an infinite Sierpinski gasket so that each edge is contained in the boundary of either the plane sector occupied by the gasket or a triangular region of the gasket's complement. If a(n) = 0, the n-th edge is contained in the sector boundary, otherwise the relevant triangular region seems to have side 2^(a(n)-1). See A307672 for a fuller description. The conjectured formulas below (that use A094373) derive from summing areas of regions within the gasket. - Corrected by Peter Munn, Aug 09 2019
From Charlie Neder, Jul 05 2019: (Start)
For each n, define the "2-balanced ternary expansion" E(n) as follows:
- E(n) begins with 0 or 1, according to the parity of n.
- The following digits are +, 0, or - as in standard balanced ternary, except + and - correspond to +2 and -2, respectively.
For example, we have E(4) = 0+-, E(7) = 10-, and E(13) = 1+-.
Then a(n) is the distance from the end of the rightmost 0, counting the last digit as 1, or 0 if 0 never appears. (End)

Examples

			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
		

Crossrefs

Sequences with similar definitions: A309054, A335933.
A055246, A191108, A306556 relate to the Cantor set.

Programs

  • PARI
    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

Formula

Alternative definition: (Start)
a(m*3^k - 3^(k-1) + A191108(i)) = k for k >= 1, 1 <= i <= 2^(k-1), all integer m.
a(A191108(i)) = a(-A191108(i)) = 0 for i >= 1.
(End)
if a(n) = k >= 1, a(3^k+n) = a(3^k-n) = k.
a(n) = a(12*3^k + n) for k >= 0, 0 <= n <= 3^k.
if a(n) = a(n') and a(n+1) = a(n'+1) then a(n*3^k + i) = a(n'*3^k + i) for k >= 0, 0 <= i <= 3^k.
a((m-1)*3^k + 1) = a((m+1)*3^k - 1) for k >= 1, all integer m.
Upper bound relations: (Start)
for k >= 2, let m_k = A034472(k-2) = 3^(k-2)+1.
a(n) < k, for -m_k < n < m_k.
a(-m_k) = a(m_k) = k.
(End)
for k>=0, a( 3^k-1) = k+1, a( 3^k+1) = k+2.
for k>=0, a(2*3^k-1) = 0, a(2*3^k+1) = k+1.
for k>=0, a(4*3^k-1) = k+1, a(4*3^k+1) = 0.
for k>=0, a(5*3^k-1) = k+3, a(5*3^k+1) = k+1.
for k>=0, a(7*3^k-1) = k+1, a(7*3^k+1) = k+3.
for k>=0, a(8*3^k-1) = k+2, a(8*3^k+1) = k+1.
A051064(i) = min{a(n) : |n-i| = 1, a(n) > 0}.
A055246(i+1) = min{n : n > A055246(i) + 1, a(n) = a(A055246(i) + 1)}.
Sum_{n=-3^k..3^k-1} A094373(a(n)) = 3 * 4^k (conjectured).
Sum_{n=-3m..3m-1} A094373(a(n)) = 4 * Sum_{n=-m..m-1} A094373(a(n)) (conjectured).
From Charlie Neder, Jul 05 2019: (Start)
Let P(n) be the power of 3 (greater than 1) closest to n and T(n) be the distance from the end - counting the last digit as 1 - of the rightmost 0 in the balanced ternary expansion of n.
If n is even, a(n) = T(n/2).
If n is odd, a(n) = T((P(n)-n)/2), or 0 if this number exceeds log_3(P(n)). (End)
Showing 1-3 of 3 results.