A321257 -id:A321257 - cp's OEIS frontend

A321237 Start with a square of dimension 1 X 1, and repeatedly append along the squares of the previous step squares with half their side length that do not overlap with any prior square; a(n) gives the number of squares appended at n-th step.

1, 8, 28, 68, 148, 308, 628, 1268, 2548, 5108, 10228, 20468, 40948, 81908, 163828, 327668, 655348, 1310708, 2621428, 5242868, 10485748, 20971508, 41943028, 83886068, 167772148, 335544308, 671088628, 1342177268, 2684354548, 5368709108, 10737418228, 21474836468

Offset: 1

Rémy Sigrist, Nov 01 2018

The following diagram depicts the first three steps of the construction:
                       +----+----+----+----+
                       |  3 |  3 |  3 |  3 |
                  +----+----+----+----+----+----+
                  |  3 |         |         |  3 |
             +----+----+    2    |    2    +----+----+
             |  3 |  3 |         |         |  3 |  3 |
        +----+----+----+---------+---------+----+----+----+
        |  3 |         |                   |         |  3 |
        +----+    2    |                   |    2    +----+
        |  3 |         |                   |         |  3 |
        +----+---------+         1         +---------+----+
        |  3 |         |                   |         |  3 |
        +----+    2    |                   |    2    +----+
        |  3 |         |                   |         |  3 |
        +----+----+----+---------+---------+----+----+----+
             |  3 |  3 |         |         |  3 |  3 |
             +----+----+    2    |    2    +----+----+
                  |  3 |         |         |  3 |
                  +----+----+----+----+----+----+
                       |  3 |  3 |  3 |  3 |
                       +----+----+----+----+
A square of step n+1 touches one or two squares of step n.
The limiting construction is an octagon (truncated square); its area is 7 times the area of the initial square.
See A321257 for a similar sequence.

Colin Barker, Table of n, a(n) for n = 1..1000
Rémy Sigrist, Illustration of the construction after 7 steps
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A154117, A321257.

a(n) = if (n==1, return (1), return (4*( 2^(n-1) + 3 * floor( (2^(n-2)-1) ) )))

Vec(x*(1 + 2*x)*(1 + 3*x) / ((1 - x)*(1 - 2*x)) + O(x^40)) \\ Colin Barker, Nov 02 2018

a(n) = 4 * (2^(n-1) + 3 * (2^(n-2)-1)) for any n > 1.
a(n) = 4 * A154117(n-1) for any n > 1.
Sum_{n > 0} a(n) / 4^(n-1) = 7.
From Colin Barker, Nov 02 2018: (Start)
G.f.: x*(1 + 2*x)*(1 + 3*x) / ((1 - x)*(1 - 2*x)).
a(n) = 5*2^n - 12 for n>1.
a(n) = 3*a(n-1) - 2*a(n-2) for n>3.
(End)

A330844 Start with a Koch snowflake tile, and repeatedly append along the tiles of the previous step tiles scaled by a factor of sqrt(1/3) and rotated by 90 degrees that do not overlap with any prior tile and of which one third of the perimeter matches one sixth of the perimeter of adjacent prior tiles; a(n) gives the number of tiles appended at n-th step.

Original entry on oeis.org

1, 6, 18, 42, 90, 186, 366, 690, 1278, 2322, 4182, 7482, 13326, 23682, 42006, 74442, 131838, 233394, 413094, 731034, 1293582, 2288898, 4049910, 7165674, 12678366, 22431954, 39688902, 70221498, 124242606, 219821730, 388929174, 688129674, 1217502846, 2154118770

Offset: 1

Rémy Sigrist, Feb 16 2020

nonn

One continuous third of the perimeter of a Koch snowflake tile at step n+1 coincides with one sixth of the perimeter of an adjacent Koch snowflake tile at step n; this is the maximum matching that can be achieved with the given scaling and rotation parameters (see illustration of two adjacent Koch snowflakes at consecutive steps in Links section).
A Koch snowflake tile at step n+1 touches one or two Koch snowflakes tile at step n.
The bounding hexagons of the Koch snowflake tiles at step n lie on a trihexagonal tiling (see representation of the bounding hexagons in Links sections; hexagons of the same color lie on the same trihexagonal tiling).
We can also compute the sequence by mean of necklaces:
- let h_2 = 0 and K_2 = (111111) (this is a necklace with six 1's),
- for n = 2, 3, ...:
      h_{n+1} = h_n + the number of occurrences of "100001" in K_n
      apply the following substitutions to K_n, in that order:
      - "100001" --> "00" (this loop will come off and form a hole),
      - "1" --> "110",
      - K_{n+1} is the necklace obtained after these substitutions,
- now we have a(n) = #K_n + 6 * h_n,
- the sequence {h_n} corresponds to the number of "holes" in the construction,
- the sequence {K_n} encodes the configuration of the tiles at n-th step (excluding holes),
- we can recover this configuration, up to some rotation, as follows:
    - start from the origin pointing to the right:
    - for m = 1..#K_n:
         - place a Koch snowflake with an arm pointing to the right,
         - move one step forward,
         - if K_n(m) = 1 then turn 60 degrees to the left,
         - otherwise turn 60 degrees to the right,
    - at the end, we are at the origin again.
Apparently, the number of holes is related to A077879 in the following manner:
- h_{n+1} - h_n = 6*A077879(n-6) for n >= 6.

Aperiodic Rhomb Tiling, Koch Tiles
Ben Van Dusen, Billy Scannel, Richard Taylor, A Fractal Comparison of Escher and Koch Tessellations, arXiv:1209.2180 [physics.pop-ph], 2012-2013.
Rémy Sigrist, Two adjacent Koch snowflakes at consecutive steps
Rémy Sigrist, Representation of the bounding hexagons for the first 10 steps
Rémy Sigrist, Representation of the figure after 13 steps
Rémy Sigrist, C# program for A330844
Rémy Sigrist, C++ program for A330844
Spacefilling Curves, On filling the Koch Snowflake
Wikipedia, Koch snowflake
Wikipedia, Trihexagonal tiling

Cf. A077879, A321257.

cp's OEIS Frontend

A321237 Start with a square of dimension 1 X 1, and repeatedly append along the squares of the previous step squares with half their side length that do not overlap with any prior square; a(n) gives the number of squares appended at n-th step.

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