Task:

Given 20 numbers that create a sequence. A few numbers
which go in a row are equal to each other. 1) Find and display the quantity
such numbers. 2) Print how many different numbers are in the sequence

Error:

for i in n: TypeError: 'int' object is not iterable

import random
numbers = []
k = 0
for i in range(20):
    n = random.randint(1, 20)
    numbers.append(n)
    print(n, end=',')
print()
for i in n:
    if n[i] == n[i-1] == n[i+1]:
        k += 1
        print(k)

Answer 1, authority 100%

Problem solution:

import random
numbers = [random.randint(1, 10) for _ in range(20)]
print(numbers)
k = 0
i = 0
while i in range(len(numbers)):
  k_ = k
  for j in range(i+1, len(numbers)):
    if numbers[i] == numbers[j]:
      k += 1
    else:
      break
  if k != k_:
    k += 1
    i += k-k_
  else:
    i += 1
print(f'Total length of all series: {k}')
print(f'Various elements: {len(set(numbers))}')

Result:

[7, 7, 1, 10, 2, 2, 7, 7, 8, 8, 2, 10, 7, 4, 4, 6, 1, 10, 8, 4]
Total length of all series: 10
Various elements: 7

Answer 2, authority 100%

I don’t understand a bit what you need, but I tried to follow your code

import random
numbers = []
k = 0
for i in range(20):
    n = random.randint(1, 20)
    numbers.append(n)
    print(n, end=',')
for i in range(len(numbers)):
    try:
        if numbers[i] == numbers[i-1] == numbers[i+1]:
            k += 1
    except IndexError:
        pass
print("\n" + str(k))