Print in ascending order the natural first n which are multiples of i or j or both

Question

Data n and two positive integers i and j different from 0, print in ascending order the first natural n s which are multiples of i or of j and or both.

Example: For n = 6 , i = 2 and j = 3 the output should be : 0,2,3,4,6,8.

n = int(input("Digite n: "))

i = int(input("Digite i: "))

j = int(input("Digite j: "))

lista = []

for k in range(i,i+1):
    for l in range(0,n):
        print("{}*{} = {}".format(k,l,k*l))
        lista.append(k*l)



for k in range(j,j+1):
    for l in range(0,n):
        print("{}*{} = {}".format(k,l,k*l))
        lista.append(k*l)


print(lista)

The above program with n =6, i =2 and j=3 is giving as output:

[0, 2, 4, 6, 8, 10, 0, 3, 6, 9, 12, 15]

For the output to be correct, I would need to order the above list in ascending order, clearing the duplicate values:

[0,2,3,4,6,8,9,10,12,15]

and then take the first 6 values, getting: [0,2,3,4,6,8], which is the expected output.

Any idea how to improve/correct the above program?

Answer 1

If all you need is a list of n numbers because the overhead of adding more and then removing them unnecessarily?

You can use a while cycle and module to see if any of the nums i and j is multiplied by each number:

n = 6
i = 2
j = 3

nat_nums, x = [], 0
while len(nat_nums) < n: # enquanto a a nossa lista tiver menos que `n` elementos
    if(x%i == 0 or x%j == 0): # modulo para ver se e multiplo de algum ou ambos
        nat_nums.append(x) # adicionar `x` a nossa lista
    x += 1

print(nat_nums) # [0, 2, 3, 4, 6, 8]

DEMONSTRATION

Using a function and a Generator:

def get_multis(n, i, j):
    x = 0
    while n > 0:
        if(x%i == 0 or x%j == 0): # modulo para ver se e multiplo de algum ou ambos
            yield x
            n -= 1
        x += 1

n, i, j = 6, 2, 3
print(list(get_multis(n, i, j))) # [0, 2, 3, 4, 6, 8]

DEMONSTRATION

PS: this way is already ordered.