A373226 Number of points in a diagonal Vicsek fractal subset of an n X n square.

0, 1, 2, 5, 6, 9, 10, 13, 18, 25, 26, 29, 30, 33, 38, 45, 46, 49, 50, 53, 58, 65, 74, 85, 90, 97, 110, 125, 126, 129, 130, 133, 138, 145, 146, 149, 150, 153, 158, 165, 174, 185, 190, 197, 210, 225, 226, 229, 230, 233

Offset

0,3

Links

Table of n, a(n) for n=0..49.

Wikipedia, Vicsek fractal.

Formula

Conjecture: a(n) = O(n^log_3(5)). This is motivated by the log_3(5) Hausdorff dimension of the diagonal Vicsek fractal.

Conjecture: a(3n) = 5*a(n). This is motivated by the recursive construction of the diagonal Vicsek fractal.

Conjecture: a(2n+1) = A151907(n)

Example

For n=3, a(3)=5 by counting x's in the following ASCII pattern:
  x.x
  .x.
  x.x
For n=4, a(4)=6 by counting x's in the following ASCII pattern:
  x...
  .x.x
  ..x.
  .x.x
For n=5, a(5)=9 by counting x's in the following ASCII pattern:
  x...x
  .x.x.
  ..x..
  .x.x.
  x...x
Generally for odd n, one can construct a diagonal Vicsek fractal on a 3^k X 3^k matrix such that 3^k >= n: place an n X n square in the center and count the x's.
For even n, there are 4 ways to most "centrally" place an n X n square; however, due to 4-fold symmetry of the diagonal Vicsek fractal they result in the same value. In our ASCII convention above we use the top-left selection.

Prog

(Python)
import copy
def combine(matrix): #accepts a matrix of matrices and fuses them
    aggregate = []
    for row in matrix:
        for j in range(0, len(matrix[0][0])):
            agg_row = []
            for block in row:
                agg_row += block[j]
            aggregate.append(agg_row)
    return aggregate
def descend(seed, zero, source, bound): #general fractal constructor
    for i in range(0, bound):
        for q in range(0, len(source)):
            for r in range(0, len(source[q])):
                if source[q][r] == 1:
                    source[q][r] = copy.deepcopy(seed)
                if source[q][r] == 0:
                    source[q][r] = copy.deepcopy(zero)
        source = combine(source) #fuse it up
    return source
def count(matrix, bound):
    counter = 0
    center_x, center_y = len(matrix)//2, len(matrix)//2
    shift_limit = bound//2
    if len(matrix)%2 == 1 and bound % 2 == 1:
        for i in range(-shift_limit, shift_limit+1):
            for j in range(-shift_limit, shift_limit+1):
                if matrix[center_x+i][center_y+j] == 1:
                    counter += 1
    if len(matrix)%2 == 1 and bound % 2 == 0:
        for i in range(-shift_limit, shift_limit):
            for j in range(-shift_limit, shift_limit):
                if matrix[center_x+i][center_y+j] == 1:
                    counter += 1
    return counter
seed = [[1, 0, 1], [0, 1, 0], [1, 0, 1]]
source = [[1]]
zero = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
#example with n=5
n=5
print(count(descend(seed, zero, source, 2), 5)) #this constructs a 3^2 X 3^2 matrix and counts the center 5 X 5 matrix

Crossrefs

Cf. A151907 (conjectured to be only the odd terms).

Sequence in context: A003664 A065890 A162177 * A213730 A191172 A078632

Adjacent sequences: A373223 A373224 A373225 * A373227 A373228 A373229

Keyword

easy,nonn

Author

Sidharth Ghoshal, May 28 2024

Status

approved