Given a series of integers from 1 to infinity and a number N.
The task is to remove every (i + 1)-th element from the remaining series at every i-th iterations and find that the given number N exists in the series or not.
Numbers in the Flavius Sieve are called Flavius Numbers.
Flavius sieve starts with the natural numbers and keep repeating the below step:
At the k-th sieving step, remove every (k+1)-st term of the sequence remaining of N natural numbers after the (k-1)-st sieving step.
For Example: 1, 3, 7, 13, 19, 27, 39, 49,
Input: N = 17
Series after i-th iterations
1). 1, 3, 5, 7, 9, 11, 13, 15, 17, …
2). 1, 3, 7, 9, 13, 15, 19, 21, 25, …
3). 1, 3, 7, 13, 15, 19, 25, …
4). 1, 3, 7, 13, 19, 27, ….
Input: N = 3
- If the given number is even so the answer is simply “No”. Because in the first iteration all the even number has been eliminated from the series.
- Repeat this process.
Else remove the number of elements removed at 1st iteration i.e. (1/2)th the number and then check
if it is divisible by 3 the answered should be “No”, else subtract the numbers before it that were
removed i.e. (1/3)rd the number and so on.
- Repeat above step for all iterations until we get answer.
- Else remove the number of elements removed at 1st iteration i.e. (1/2)th the number and then check
Below is the implementation of the approach:
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