in finding the Fabricated from Consecutive Fib Numbers in Python

0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6, 5, 4, 0, 5, 3, 0, 3, …

That is the Van Eck’s Series.

Let’s undergo it step-by-step.

Time period 1: The primary time period is 0.
Time period 2: Since we haven’t observed 0 ahead of, the second one time period is 0.
Time period 3: Since we had observed a nil ahead of, one step again, the 3rd time period is 1
Time period 4: Since we haven’t observed a 1 ahead of, the fourth time period is 0
Time period 5: Since we had observed a nil ahead of, two steps again, the 5th time period is two.
And so forth…

Your activity is to search out the n_th quantity in Van Eck’s Series. (1-based)

The Answer in Python

Choice 1

from collections import Counter

c=Counter()
SEQ = [0]
for i in vary(1000):
    n = SEQ[-1]
    if no longer c[n]: c[n]=i
    SEQ.append(i-c[n])
    c[n]=i
    
seq=SEQ.__getitem__

Choice 2

def dist(arr):
    for i in vary (1, len(arr)):
        if arr[-1-i] == arr[-1]:
            go back i
    go back 0

def seq(n):
    s = [0, 0]
    for _ in vary (n):
        s.append(dist(s))
    go back s[n-1]

def seq(n):
    van, eck = [0], 0
    whilst n := n - 1:
        van.insert(0, eck := van.index(eck, 1) if eck in van[1:] else 0)
    go back eck

Check instances to validate the answer

from answer import seq
import check

from random import randint

@check.describe("Pattern assessments:")
def assessments():
    @check.it("Small numbers")
    def _():
        s = [0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6]
        for i in vary (len(s)):
            check.assert_equals(seq(i+1), s[i])
    @check.it('Greater numbers')
    def __():
        s = [3, 1, 42, 0, 5, 15, 20, 0, 4, 32, 0, 3, 11,
             18, 0, 4, 7, 0, 3, 7, 3, 2, 31, 0, 6, 31, 3,
             6, 3, 2, 8, 33, 0, 9, 56, 0, 3, 8, 7, 19, 0,
             5, 37, 0, 3, 8, 8, 1, 46, 0, 6, 23, 0]
        for i in vary (len(s)):
            check.assert_equals(seq(i+50), s[i])

@check.describe('Random assessments:')
def r():

    def dist(arr):
        for i in vary (1, len(arr)):
            if arr[-1-i] == arr[-1]:
                go back i
        go back 0

    def ref_sol(n):
        s = [0, 0]
        for _ in vary (n):
            s.append(dist(s))
        go back s[n-1]
    
    @check.it('200 random assessments:')
    def _():
        for _ in vary (200):
            a = randint(100, 1000)
            exp = ref_sol(a)
            check.assert_equals(seq(a), exp)