count pattern.ipynb

Kernel: Python 3 (system-wide)

Counting Pattern

Define $\mathcal{C}(n)$ as the recursively defined sequence of integers whose $i$th term is the 'verbal summary' of the digits in the previous term and whose first term is $n$.

Example:

If $1232$ is a term, the next term is $112213$ because $1232$ contains one $1$, two $2$s, and one $3$.

import numpy as np
from numba import njit
import matplotlib.pyplot as plt

@njit
def split(it:str):
    return [char for char in it]

def nextNum(initNum:str):
    splitUp = split(initNum)
    seq = ''
    for occurance in range(max([int(i) for i in splitUp])+1):
        howMany = splitUp.count(str(occurance))
        if howMany != 0:
            seq += str(howMany) + str(occurance)
    return seq
#     return ''.join(str(splitUp.count(str(occurance))) + str(occurance) if splitUp.count(str(occurance)) != 0 else '' for occurance in range(max([int(i) for i in splitUp])+1))

def C(initialNumber:str):
    tot = [initialNumber]
    prev = initialNumber
    while True:
        nextNumber = nextNum(prev)
        if nextNumber == prev:
            break
        tot += [nextNumber]
        prev = nextNum(prev)
    return tot, len(tot)

test a bunch

data = np.zeros(50)

for a in range(1,50):
    data[a] = C(str(a))[1]

C('39')

(['39',
  '1319',
  '211319',
  '31121319',
  '41122319',
  '3122131419',
  '4122231419',
  '3132132419',
  '3122331419'],
 9)

data[40]

0.0

    fig, ax = plt.subplots(figsize=(7,7))
    ax.set_aspect('equal')
    ax.autoscale()

    x=np.linspace(-1,1,500)
    y_1 = np.sqrt(1-x**2)
    y_2 = -np.sqrt(1-x**2)

    ax.plot(x,y_1,'k');
    ax.plot(x,y_2,'k');
    ax.plot(*generate(0,angle,bounces));