Trying to find the next prime number

Question

MyFunctions file file -

def factList(p,n1):
    counter = 1
    while counter <= n1:
        if n1 % counter == 0:
            p.append(counter)
        counter = counter + 1    



def isPrime(lst1,nbr):
    factList(lst1, nbr)
    if len(lst1) == 2:
        return True
    else:
        return False





def nextPrime(nbr1):
    cnt1 = 1
    while cnt1 == 1:
        nbr1 == nbr1 + 1
        if isPrime(lst2,nbr1):
            cnt1 = 0

Filetester file -

nbr1 = 13 
nextPrime(nbr1)
print nbr1

My isPrime function already works I'm tring to use my isPrime function for my nextPrime function, when I run this I get ">>> 13

" (when using 13)

">>> " (When using 14)

I am supposed to get 17 not 13. And if I change it to a composite number in function tester it gets back in a infinite loop. Please only use simple functions (the ones I have used in my code).

Answer 1

This is NOT the right way to do this, but this is the closest adaptation of your code that I could do:

def list_factors_pythonic(number):
    """For a given number, return a list of factors."""
    factors = []
    for x in range(1, number + 1):
        if number % x == 0:
            factors.append(x)
    return factors

def list_factors(number):
    """Alternate list_factors implementation."""
    factors = []
    counter = 1
    while counter <= number:
        if number % counter == 0:
            factors.append(counter)
    return factors  

def is_prime(number):
    """Return true if the number is a prime, else false."""
    return len(list_factors(number)) == 2

def next_prime(number):
    """Return the next prime."""
    next_number = number + 1
    while not is_prime(next_number):
        next_number += 1
    return next_number

Answer 2

This would be helpful:

def nextPrime(number):
    for i in range(2,number):
        if number%i == 0:
            return False
        sqr=i*i
        if sqr>number:
           break
    return True

number = int(input("Enter the num: ")) + 1
while(True):
    res=nextPrime(number)
    if res:
        print("The next number number is: ",number)
        break
    number += 1

Answer 3

I don't know python but if it's anything like C then you are not assigning anything to your variables, merely testing for equality.

while cnt1 == 1:
    nbr1 == nbr1 + 1
    if isPrime(lst2,nbr1):
        cnt1 == cnt1 + 1

Should become

while cnt1 == 1:
    nbr1 = nbr1 + 1             << changed here
    if isPrime(lst2,nbr1):
        cnt1 = cnt1 + 1          << and here

Answer 4

Well this code help you

n=int(input())
p=n+1
while(p>n):
    c=0
    for i in range(2,p):
        if(p%i==0):
            break
        else:c+=1
    if(c>=p-2):
        print(p)
        break
    p+=1

Answer 5

this code optimized for finding sudden next prime number of a given number.it takes about 6.750761032104492 seconds

def k(x):
   return pow(2,x-1,x)==1
n=int(input())+1
while(1):
    if k(n)==True:
        print(n)
        break
n=n+1