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A385414 Number of distinct states of Conway's Game of Life, starting from an n-th level Hilbert curve on a toroidal 2^(n+1)-1 by 2^(n+1)-1 grid.

2, 2, 3, 24, 70, 584, 1325, 2082, 5304, 6327, 10679, 11822

Offset: 0

Luke Bennet, Jun 27 2025

The curve is taken with segments of length 2 so that it follows a path through coordinates (2*A059252(t), 2*A059253(t)) for 0 <= t < 2^n.
The size of the toroidal grid is the extent of these coordinates so that the cells on one edge are immediately adjacent to the cells on the opposite side.
The grid has a fixed position and orientation and states differing at any cell are distinct.

			For n=0, the curve is a single cell on a 1 X 1 toroidal grid and has a(0) = 2 states: initially live, then dead and remaining so.
For n=2 the initial state and two subsequent states are
  o o o . o o o | . . . . . . . | . . . . . . . |
  o . o . o . o | . . . . . . . | . . . . . . . |
  o . o o o . o | . . o . o . . | . . . . . . . |
  o . . . . . o | . . . . . . . | . . . . . . . |
  o o o . o o o | . . o . o . . | . . . . . . . |
  . . o . o . . | . . . . . . . | . . . . . . . |
  o o o . o o o | . . . . . . . | . . . . . . . |
  (generation 1)  (generation 2)  (generation 3)
Every generation after 3 is identical to generation 3, so this sequence has 3 distinct states. Thus, a(2) = 3.

Cf. A059252, A059253.

A385493 Number of distinct states in Conway's Game of Life acting on a (2n+1) X (2n+1) toroidal grid starting with (x,y) turned on if and only if x-n + (y-n)*i is a Gaussian prime.

Original entry on oeis.org

1, 1, 1, 6, 9, 4, 4, 5, 14, 12, 17, 5, 8, 19, 15, 34, 20, 21, 19, 77, 52, 29, 58, 39, 27, 27, 68, 31, 27, 27, 27, 70, 49, 129, 83, 43, 153, 40, 82, 128, 60, 457, 436, 79, 99, 71, 71, 178, 125, 281, 121, 121, 94, 231, 94, 94, 385, 122, 94, 94, 175, 306, 156

Offset: 0

Luke Bennet, Jun 30 2025

nonn

			For a(3), the sequence of Conway's Game of Life is
 | . o . o . o . | . o . o . o . | . o . . . o . |
 | o . o . o . o | o . . . . . o | o o o o o o o |
 | . o o . o o . | . . o . o . . | . o . . . o . |
 | o . . . . . o | o . . . . . o | . o . . . o . |
 | . o o . o o . | . . o . o . . | . o . . . o . |
 | o . o . o . o | o . . . . . o | o o o o o o o |
 | . o . o . o . | . o . o . o . | . o . . . o . |
  (generation 1)  (generation 2)  (generation 3)
 | . . . o . . . | . . o o o . . | . o . . . o . |
 | . . . o . . . | . . o o o . . | o . . . . . o |
 | . . . o . . . | o o . o . o o | . . . . . . . |
 | o o o . o o o | o o o . o o o | . . . . . . . |
 | . . . o . . . | o o . o . o o | . . . . . . . |
 | . . . o . . . | . . o o o . . | o . . . . . o |
 | . . . o . . . | . . o o o . . | . o . . . o . |
  (generation 4)  (generation 5)  (generation 6)
Every generation after 6 is identical to generation 6, so this sequence has 6 unique states. Thus, a(3) = 6.

Cf. A055025.

import torch
import numpy as np
def prime_mask(limit):
    is_prime = torch.ones(limit + 1, dtype=torch.bool)
    is_prime[:2] = False
    for i in range(2, int(limit**0.5) + 1):
        if is_prime[i]:
            is_prime[i*i : limit+1 : i] = False
    return is_prime
def Gauss_primes(N):
    A, B = torch.meshgrid(torch.arange(-N, N+1), torch.arange(-N, N+1))
    norm = A**2 + B**2
    is_prime = prime_mask(2*N**2)
    mask = (A != 0) & (B != 0) & is_prime[norm]
    axis_mask = ((A == 0) ^ (B == 0))
    axis_val = (A + B).abs()
    axis_mask &= is_prime[axis_val] & ((axis_val % 4) == 3)
    return mask | axis_mask
def update(G):
    shifts = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
    neighbors = sum(torch.roll(G, shifts=shift, dims=(0,1)) for shift in shifts)
    return (G & ((neighbors == 2) | (neighbors == 3))) | (~G & (neighbors == 3))
def a(n):
    if n == 0 or n == 1:
        return 1
    G = Gauss_primes(n).to("cuda").to(torch.uint8)
    seen, step = set(), 0
    while True:
        flat = G.flatten().to("cpu").numpy()
        key = bytes(np.packbits(flat))
        if key in seen:
            return step
        seen.add(key)
        G = update(G)
        step += 1

A384509 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = ((5/n+1) + (5/n-1)i)/(nsqrt(2)) + 1/4 + (1/2)*i to reach |z| > 2, starting with z = 0.

Original entry on oeis.org

1, 2, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 51, 51, 53, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 71, 71, 73, 74, 75, 76, 77, 78, 79, 80

Offset: 1

Luke Bennet, May 31 2025

nonn

a(n)/n seems to converge to Pi/(2*sqrt(2)).

a(n) counts the escape time of points outside the Mandelbrot set that converge to the Mandelbrot set's 1/4 period bulb.

Luke Bennet, Table of n, a(n) for n = 1..10001
Thies Brockmöller, Oscar Scherz, and Nedim Srkalović, Pi in the Mandelbrot set everywhere, arXiv preprint arXiv:2505.07138 [math.DS], 2025.
Aaron Klebanoff, Pi in the Mandelbrot Set, Fractals 9 (2001), nr. 4, p. 393-402.

Cf. A093954, A097486, A383750, A384513.

c(n) = ((5/n+1) + (5/n-1)*I)/(n*sqrt(2)) + 1/4 + (1/2)*I;
a(n) = my(z=0, k=0, c=c(n)); while(norml2(z)<=4, z = z^2 + c; k++); k; \\ Michel Marcus, Jun 01 2025

import mpmath
from mpmath import iv
def a(n):
    dps = 1
    while True:
        mpmath.iv.dps = dps
        c = iv.mpc(iv.mpf(5) / n + 1, iv.mpf(5) / n - 1)
        c = c / (n * iv.sqrt(2)) + 0.25 + 0.5j
        z = iv.mpc(0, 0)
        counter = 0
        while (z.real**2 + z.imag**2).b <= 4:
            z = z ** 2 + c
            counter += 1
        if (z.real**2 + z.imag**2).a > 4:
            return counter
        dps *= 2

A384513 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = 16/(n^2) + (1/n)i + 3/8 + (sqrt(3)/8)i to reach |z| > 2, starting with z = 0.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 5, 6, 7, 7, 8, 8, 9, 9, 10, 10, 10, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 27, 27, 28, 28, 28, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40

Offset: 1

Luke Bennet, May 31 2025

nonn

a(n)/n seems to converge to Pi/6.

a(n) counts the escape time of points outside the Mandelbrot set that converge to the Mandelbrot set's 1/6 period bulb. This is a proven fact and was the motivation for creating the sequence.

Luke Bennet, Table of n, a(n) for n = 1..10001
Thies Brockmöller, Oscar Scherz, and Nedim Srkalović, Pi in the Mandelbrot set everywhere, arXiv preprint arXiv:2505.07138 [math.DS], 2025.
Aaron Klebanoff, Pi in the Mandelbrot Set, Fractals 9 (2001), nr. 4, p. 393-402.

Cf. A019673, A097486, A383750, A384509.

def a(n):
    dps = 1
    while True:
        mpmath.iv.dps = dps
        c = iv.mpc(iv.mpf(16) / n ** 2 + 0.375, iv.mpf(1) / n + iv.sqrt(3) / 8)
        z = iv.mpc(0, 0)
        counter = 0
        while (z.real**2 + z.imag**2).b <= 4:
            z = z ** 2 + c
            counter += 1
        if (z.real**2 + z.imag**2).a > 4:
            return counter
        dps *= 2

A383750 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = 1/n + (2/(n^2))i - 1/8 + (3sqrt(3)/8)*i to reach |z| > 2, starting with z = 0.

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 11, 13, 15, 17, 19, 20, 22, 24, 26, 28, 29, 31, 33, 35, 37, 38, 40, 42, 44, 46, 47, 49, 51, 53, 55, 57, 58, 60, 62, 64, 66, 68, 69, 71, 73, 75, 77, 78, 80, 82, 84, 86, 87, 89, 91, 93, 95, 96, 98, 100, 102, 104, 105, 107, 109, 111, 113, 115, 116, 118, 120

Offset: 1

Luke Bennet, May 08 2025

nonn

a(n)/n appears to converge to Pi/sqrt(3).

a(n) counts the escape time of points outside the Mandelbrot set that converge to the Mandelbrot set's 1/3 period bulb.

Luke Bennet, Table of n, a(n) for n = 1..10001
Thies Brockmöller, Oscar Scherz, and Nedim Srkalović, Pi in the Mandelbrot set everywhere, arXiv preprint arXiv:2505.07138 [math.DS], 2025.
Aaron Klebanoff, Pi in the Mandelbrot Set, Fractals 9 (2001), nr. 4, p. 393-402.

Cf. A093602, A097486, A384509, A384513

import mpmath
from mpmath import iv
def a(n):
    dps = 1
    while True:
        mpmath.iv.dps = dps
        real_part = iv.mpf(1) / n - iv.mpf('0.125')
        imag_part = iv.mpf(2) / (n ** 2) + 3 * iv.sqrt(3) / 8
        c = iv.mpc(real_part, imag_part)
        z = iv.mpc(0, 0)
        counter = 0
        while (z.real**2 + z.imag**2).b <= 4:
            z = z ** 2 + c
            counter += 1
        if (z.real**2 + z.imag**2).a > 4:
            return counter
        dps *= 2

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User: Luke Bennet

A385414 Number of distinct states of Conway's Game of Life, starting from an n-th level Hilbert curve on a toroidal 2^(n+1)-1 by 2^(n+1)-1 grid.

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A385493 Number of distinct states in Conway's Game of Life acting on a (2n+1) X (2n+1) toroidal grid starting with (x,y) turned on if and only if x-n + (y-n)*i is a Gaussian prime.

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A384509 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = ((5/n+1) + (5/n-1)*i)/(n*sqrt(2)) + 1/4 + (1/2)*i to reach |z| > 2, starting with z = 0.

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A384513 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = 16/(n^2) + (1/n)*i + 3/8 + (sqrt(3)/8)*i to reach |z| > 2, starting with z = 0.

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A383750 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = 1/n + (2/(n^2))*i - 1/8 + (3*sqrt(3)/8)*i to reach |z| > 2, starting with z = 0.

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Python

A384509 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = ((5/n+1) + (5/n-1)i)/(nsqrt(2)) + 1/4 + (1/2)*i to reach |z| > 2, starting with z = 0.

A384513 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = 16/(n^2) + (1/n)i + 3/8 + (sqrt(3)/8)i to reach |z| > 2, starting with z = 0.

A383750 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = 1/n + (2/(n^2))i - 1/8 + (3sqrt(3)/8)*i to reach |z| > 2, starting with z = 0.