Our function won't be very interesting until it approaches 100, so let's find out when it does this.

f(1) = 50
For a while, f(n) = 50 + 3(n-1)
So,
f(17) = 98
f(18) = 101
f(19) = 99
f(20) = 102
f(21) = 100
f(22) = 103
f(23) = 101

Now, a pattern has been established: f(a+5k) = f(a) while a>=18 and k is a whole number.

f(500) = f(20+5(96)) = f(20) = 102