

The Padovan sequence is similar to the Fibonacci sequence in several ways.
Some are given in the table below, and the referenced video shows some of the geometric
similarities.



 Comment 
 Padovan 
 Fibonacci


 
 



 Named after. 
 Richard Padovan 
 Leonardo of Pisa: Fibonacci


 
 



 Recurrence initial values. 
 P(0)=P(1)=P(2)=1 
 F(0)=0, F(1)=1


 Recurrence relation. 
 P(n)=P(n-2)+P(n-3) 
 F(n)=F(n-1)+F(n-2)


 
 



 First 10 terms. 
 1,1,1,2,2,3,4,5,7,9 
 0,1,1,2,3,5,8,13,21,34


 
 



 Ratio of successive terms... 
 Plastic ratio, p 
 Golden ratio, g


 
 1.324717957244746025960908854… 
 1.6180339887498948482...


 Exact formula of ratios p and q. 
 ((9+69**.5)/18)**(1/3) + ((9-69**.5)/18)**(1/3) 
 (1+5**0.5)/2


 Ratio is real root of polynomial. 
 p: x**3-x-1 
 g: x**2-x-1


 
 



 Spirally tiling the plane using. 
 Equilateral triangles 
 Squares


 
 



 Constants for ... 
 s= 1.0453567932525329623 
 a=5**0.5


 ... Computing by truncation. 
 P(n)=floor(p**(n-1) / s + .5) 
 F(n)=floor(g**n / a + .5)


 
 



 L-System Variables. 
 A,B,C 
 A,B


 L-System Start/Axiom. 
 A 
 A


 L-System Rules. 
 A->B,B->C,C->AB 
 A->B,B->AB

Task

 Write a function/method/subroutine to compute successive members of the Padovan series using the recurrence relation.
 Write a function/method/subroutine to compute successive members of the Padovan series using the floor function.
 Show the first twenty terms of the sequence.
 Confirm that the recurrence and floor based functions give the same results for 64 terms,
 Write a function/method/... using the L-system to generate successive strings.
 Show the first 10 strings produced from the L-system
 Confirm that the length of the first 32 strings produced is the Padovan sequence.
Show output here, on this page.

Ref

 The Plastic Ratio - Numberphile video.


from math import floor
from collections import deque
from typing import Dict, Generator
 
 
def padovan_r() -> Generator[int, None, None]:
    last = deque([1, 1, 1], 4)
    while True:
        last.append(last[-2] + last[-3])
        yield last.popleft()
 
_p, _s = 1.324717957244746025960908854, 1.0453567932525329623
 
def padovan_f(n: int) -> int:
    return floor(_p**(n-1) / _s + .5)
 
def padovan_l(start: str='A',
             rules: Dict[str, str]=dict(A='B', B='C', C='AB')
             ) -> Generator[str, None, None]:
    axiom = start
    while True:
        yield axiom
        axiom = ''.join(rules[ch] for ch in axiom)
 
 
if __name__ == "__main__":
    from itertools import islice
 
    print("The first twenty terms of the sequence.")
    print(str([padovan_f(n) for n in range(20)])[1:-1])
 
    r_generator = padovan_r()
    if all(next(r_generator) == padovan_f(n) for n in range(64)):
        print("\nThe recurrence and floor based algorithms match to n=63 .")
    else:
        print("\nThe recurrence and floor based algorithms DIFFER!")
 
    print("\nThe first 10 L-system string-lengths and strings")
    l_generator = padovan_l(start='A', rules=dict(A='B', B='C', C='AB'))
    print('\n'.join(f"  {len(string):3} {repr(string)}"
                    for string in islice(l_generator, 10)))
 
    r_generator = padovan_r()
    l_generator = padovan_l(start='A', rules=dict(A='B', B='C', C='AB'))
    if all(len(next(l_generator)) == padovan_f(n) == next(r_generator)
           for n in range(32)):
        print("\nThe L-system, recurrence and floor based algorithms match to n=31 .")
    else:
        print("\nThe L-system, recurrence and floor based algorithms DIFFER!")