Python| sympy.fibonacci() 方法

借助sympy.fibonacci()方法，我们可以在 SymPy 中找到斐波那契数和斐波那契多项式。

斐波那契(n) -

斐波那契数是由初始项定义的整数序列$F_0 = 0$ , $F_1 = 1$和两项递推关系$F_n = F_{n-1} + F_{n-2}$ .

Syntax: fibonacci(n)

Parameter:
n – It denotes the number upto which Fibonacci number is to be calculated.

Returns: Returns the n^th Fibonacci number.

示例 #1：

# import sympy 
from sympy import * 
  
n = 7
print("Value of n = {}".format(n))
   
# Use sympy.fibonacci() method 
nth_fibonacci = fibonacci(n)  
      
print("Value of nth fibonacci number : {}".format(nth_fibonacci))  

输出：

Value of n = 7
Value of nth fibonacci number : 13

斐波那契(n, k) -

斐波那契多项式定义为$F_1(k) = 1$ , $F_2(k) = k$ ， 和$F_n(k) = k*F_{n-1}(k) + F_{n-2}(k)$为了$n > 2$ .对于所有正整数$n$ , $F_n(1) = F_n$ .

Syntax: fibonacci(n, k)

Parameter:
n – It denotes the n^th Fibonacci polynomial.
k – It denotes the variable in the Fibonacci polynomial.

Returns: Returns the the nth Fibonacci polynomial in k, F_n(k)

示例 #2：

# import sympy 
from sympy import * 
  
n = 5
k = symbols('x')
print("Value of n = {} and k = {}".format(n, k))
   
# Use sympy.fibonacci() method 
nth_fibonacci_poly = fibonacci(n, k)  
      
print("The nth fibonacci polynomial : {}".format(nth_fibonacci_poly))  

输出：

Value of n = 5 and k = x
The nth fibonacci polynomial : x**4 + 3*x**2 + 1

示例#3：

# import sympy 
from sympy import * 
  
n = 6
k = 3
print("Value of n = {} and k = {}".format(n, k))
   
# Use sympy.fibonacci() method 
nth_fibonacci_poly = fibonacci(n, k)  
      
print("The nth fibonacci polynomial value : {}".format(nth_fibonacci_poly))  

输出：

Value of n = 6 and k = 3
The nth fibonacci polynomial value : 360