Unlucky number 13

Question

I came across this problem Unlucky number 13! recently but could not think of efficient solution this.

Problem statement :

N is taken as input.

Answer 1

Let f(n) be the number of sequences of length n that have no "13" in them, and g(n) be the number of sequences of length n that have "13" in them.

Then f(n) = 10^n - g(n) (in mathematical notation), because it's the number of possible sequences (10^n) minus the ones that contain "13".

Base cases:

f(0) = 1
g(0) = 0
f(1) = 10
g(1) = 0

When looking for the sequences with "13", a sequence can have a "13" at the beginning. That will account for 10^(n-2) possible sequences with "13" in them. It could also have a "13" in the second position, again accounting for 10^(n-2) possible sequences. But if it has a "13" in the third position, and we'd assume there would also be 10^(n-2) possible sequences, we could those twice that already had a "13" in the first position. So we have to substract them. Instead, we count 10^(n-4) times f(2) (because those are exactly the combinations in the first two positions that don't have "13" in them).

E.g. for g(5):

g(5) = 10^(n-2) + 10^(n-2) + f(2)*10^(n-4) + f(3)*10^(n-5)

We can rewrite that to look the same everywhere:

g(5) = f(0)*10^(n-2) + f(1)*10^(n-3) + f(2)*10^(n-4) + f(3)*10^(n-5)

Or simply the sum of f(i)*10^(n-(i+2)) with i ranging from 0 to n-2.

In Python:

from functools import lru_cache

@lru_cache(maxsize=1024)
def f(n):
    return 10**n - g(n)

@lru_cache(maxsize=1024)
def g(n):
    return sum(f(i)*10**(n-(i+2)) for i in range(n-1))  # range is exclusive

The lru_cache is optional, but often a good idea when working with recursion.

---

>>> [f(n) for n in range(10)]
[1, 10, 99, 980, 9701, 96030, 950599, 9409960, 93149001, 922080050]

The results are instant and it works for very large numbers.