Find all values in a list that are present in the following sequence : f(0) = 0, f(1) = 1, f(n) = 5*f(n-1) - 2*f(n-2) for all n > 1

recurrence.py

from math import ceil, log, sqrt
"""
This solution uses the closed form expression of the given linear recurrence relation. Some approximations has been made to calculate the nearest value of n. Unlike other solutions that use recursion or memoization techniques, this is a much more faster and efficient constant time O(1) solution (assuming real-valued arithmetic is constant time).
"""
phi = 4.561552813
psi = 0.4384471872
sq = sqrt(17)

def isPresent(k):
    
    if k == 0:
        return True
    
    sq = sqrt(17)
    n = log(sq*k, phi)
    m = ceil(n)
    f_m = (phi**m - psi**m)//sq
    return f_m == k

def is_part_of_series(lst):
    
    res = []
    for k in lst:
        if isPresent(k):
            res.append(k)
    print(res)

# is_part_of_series([0,1,2,3,5,23,54,105,200,3000])