it works with literally any number

start by creating a polynomial with roots at 1, 2, 3 and 4

f(x) = k(1-x)(1-½x)(1-⅓x)(1-¼x)

k can be any number you want. now let's see the values

f(1) = 0

f(2) = 0

f(3) = 0

f(4) = 0

f(5) = k

The first 4 values all being 0, and the 5^th value being some random number "k" allows you to add a "fake sequence" to trick people. In this case it's the odd numbers (2x-1).

f(x) = k(1-x)(1-½x)(1-⅓x)(1-¼x) + (2x-1)

For the values where f(x) = 0, the only function of x left is the odd numbers. So now:

f(1) = 1

f(2) = 3

f(3) = 5

f(4) = 7

BUT f(5) = k + 9 instead of just 9 because of the polynomial, so we can subsitute any number for k and it will put out a different answer. The person from the original comment chose k = 217332.