內部帶有函數的遞歸公式

Question

我想以我不擅長編碼的事實作為開頭。

我的問題是我想用遞歸公式中的函數（在python中）生成一個序列的新值。

序列由這個公式給出，我在圖片 p 中定義了必要的函數：

import math
def p (k, q):
    if math.gcd(k, q) == q:
         return 1
    elif math.gcd(k, q) == 1:
         return q 
    else:
        return alpha 

上面代碼塊中的“alpha”只是整個程序早期部分產生的一些整數。 現在，我的問題是如何為給定的整數 n 生成其對應的值，由公式給出，其中序列的前三個值是

k_1 = 1, k_2 = 2, and k_3 = 6.

---

編輯：背景：

import math
import numpy as np
# Find's the prime factorization of any given integer n,
# where the list depicts the necessary multiples to determine the factorization.
def prime_factors(q):
    i = 2
    factors = []
    while i * i <= q:
        if q % i:
            i += 1
        else:
            q //= i
            factors.append(i)
    if q > 1:
        factors.append(q)
    return factors

q = 6

pf = prime_factors(q)
print("The prime factorization of q is: " + str(pf))

# Reducing the prime factorization of q to only considering its prime factors and "getting rid of" the exponents
res = []
for i in pf:
    if i not in res:
        res.append(i)
alpha = np.prod(np.array(res))

print("The result of applying the function α is: " + str(alpha))

#Defining the function p
def p (k, q):
    global alpha
    if math.gcd(k, q) == q:
         return 1
    elif math.gcd(k, q) == 1:
         return q 
    else:
        return alpha   

# Don't know how to produce the following k_q depending on a choice of q
print("The number at the last of the following list is k_n: [?] ") 

# The list should look like: k_1 = 1, k_2 = 2, k_3 = 6, k_4 = 12, k_5 = 60, k_6 = 60, k_7 = 420, ... and so on

>> The prime factorization of q is: [2, 3]
>> The result of applying the function α is: 6
>> The number at the last of the following list is k_q: [?] 

Answer 1

我認為一個 for 循環以及一個計算系列中下一個數字的函數就足夠了。 我假設alpha = 1

我在該系列的前 6 個術語中得到的輸出是：

[1, 2, 6, 6, 30]

---

代碼：

import math

alpha = 1
def p (k, q):
    global alpha
    if math.gcd(k, q) == q:
        return 1
    elif math.gcd(k, q) == 1:
        return q
    else:
        return alpha

def get_next_num(k_prev, n):
    k_next = k_prev*p(k_prev, n)
    return k_next

k_prev = 1 #Since you said the first term of the series is 1
arr = []
for n in range(1, 6):
    k_new = get_next_num(k_prev, n)
    arr.append(k_new)
    k_prev=k_new

print(arr)