NumPy – 使用比奈公式的斐波那契数列

我们都熟悉斐波那契数列。序列中的每个数字都是它前面的两个数字的总和。所以，序列是：0, 1, 1, 2, 3, 5, 8, 13, 21, 34…… 在本教程中，我们将使用 NumPy 并借助Binet 公式来实现相同的结果。

比奈公式
$fn = \left [\left (\frac{1 + \sqrt{5}}{2} \right )^{n} - \left (\frac{1 - \sqrt{5}}{2} \right )^{n} \right ]\ast \frac{1}{\sqrt{5}} where, alpha = \left (\frac{1 + \sqrt{5}}{2} \right ), beta = \left (\frac{1 - \sqrt{5} }{2}\right )$

“ n ”是与斐波那契数列的前“n”个数字相关的参数。在第一个示例中，我们将找出斐波那契数列的前 10 个数字（n = 10），然后我们从用户那里获取参数“n”并产生相应的结果。

注意：我们忽略了斐波那契数列的第一个元素（0）

示例 1：查找前 10 个斐波那契数。

import numpy as np
   
# We are creating an array contains n = 10 elements
# for getting first 10 Fibonacci numbers
a = np.arange(1, 11)
lengthA = len(a)
  
# splitting of terms for easiness
sqrtFive = np.sqrt(5)
alpha = (1 + sqrtFive) / 2
beta = (1 - sqrtFive) / 2
  
# Implementation of formula
# np.rint is used for rounding off to integer
Fn = np.rint(((alpha ** a) - (beta ** a)) / (sqrtFive))
print("The first {} numbers of Fibonacci series are {} . ".format(lengthA, Fn))

输出 ：

The first 10 numbers of Fibonacci series are [ 1. 1. 2. 3. 5. 8. 13. 21. 34. 55.] .

示例 2：查找第一个 'n' 斐波那契数..

import numpy as np
  
# We are creating an array contains n elements
# for getting first 'n' Fibonacci numbers
fNumber = int(input("Enter the value of n + 1'th number : "))
a = np.arange(1, fNumber)
length_a = len(a)
  
# splitting of terms for easiness
sqrt_five = np.sqrt(5)
alpha = (1 + sqrt_five) / 2
beta = (1 - sqrt_five) / 2
  
# Implementation of formula
# np.rint is used for rounding off to integer
Fn = np.rint(((alpha ** a) - (beta ** a)) / (sqrt_five))
print("The first {} numbers of Fibonacci series are {} . ".format(length_a, Fn))

输出 ：

# Here user input was 10
Enter the value of n+1'th number :10
The first 9 numbers of Fibonacci series are [ 1.  1.  2.  3.  5.  8. 13. 21. 34.] .