斐波那契数列中数字的第 n 个倍数的Python程序

给定两个整数 n 和 k。在斐波那契数列中找到 K 的第 n 个倍数。
例子：

Input : k = 2, n = 3
Output : 9
3\'rd multiple of 2 in Fibonacci Series is 34 
which appears at position 9.

Input  : k = 4, n = 5 
Output : 30
5\'th multiple of 5 in Fibonacci Series is 832040 
which appears at position 30.

一个有效的解决方案基于以下有趣的属性。
斐波那契数列在模表示下总是周期性的。以下是示例。

F (mod 2) = 1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,
            1,1,0,1,1,0,1,1,0,1,1,0,1,1,0 
Here 0 is repeating at every 3rd index and 
the cycle repeats at every 3rd index. 

F (mod 3) = 1,1,2,0,2,2,1,0,1,1,2,0,2,2,1,0
            ,1,1,2,0,2,2,1,0,1,1,2,0,2,2
Here 0 is repeating at every 4th index and 
the cycle repeats at every 8th index.

F (mod 4) = 1,1,2,3,1,0,1,1,2,3,1,0,1,1,2,3,
           1,0,1,1,2,3,1,0,1,1,2,3,1,0 
Here 0 is repeating at every 6th index and 
the cycle repeats at every 6th index.

F (mod 5) = 1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,
            2,2,4,1,0,1,1,2,3,0,3,3,1,4,0
Here 0 is repeating at every 5th index and
the cycle repeats at every 20th index.

F (mod 6) = 1,1,2,3,5,2,1,3,4,1,5,0,5,5,4,
            3,1,4,5,3,2,5,1,0,1,1,2,3,5,2
Here 0 is repeating at every 12th index and 
the cycle repeats at every 24th index.

F (mod 7) = 1,1,2,3,5,1,6,0,6,6,5,4,2,6,1,
            0,1,1,2,3,5,1,6,0,6,6,5,4,2,6 
Here 0 is repeating at every 8th index and 
the cycle repeats at every 16th index.

F (mod 8) = 1,1,2,3,5,0,5,5,2,7,1,0,1,1,2,
            3,5,0,5,5,2,7,1,0,1,1,2,3,5,0 
Here 0 is repeating at every 6th index and 
the cycle repeats at every 12th index.

F (mod 9) = 1,1,2,3,5,8,4,3,7,1,8,0,8,8,7,
            6,4,1,5,6,2,8,1,0,1,1,2,3,5,8 
Here 0 is repeating at every 12th index and 
the cycle repeats at every 24th index.

F (mod 10) = 1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,
             7,7,4,1,5,6,1,7,8,5,3,8,1,9,0.
Here 0 is repeating at every 15th index and
the cycle repeats at every 60th index.

Python3

# Python Program to find position of n\'th multiple
# of a number k in Fibonacci Series
 
def findPosition(k, n):
    f1 = 0
    f2 = 1
    i =2;
    while i!=0:
        f3 = f1 + f2;
        f1 = f2;
        f2 = f3;
 
        if f2%k == 0:
            return n*i
 
        i+=1
         
    return
 
 
# Multiple no.
n = 5;
# Number of whose multiple we are finding
k = 4;
 
print("Position of n\'th multiple of k in"
                "Fibonacci Series is", findPosition(k,n));
 
# Code contributed by Mohit Gupta_OMG

输出：

Position of n'th multiple of k inFibonacci Series is 30

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