To show that every series has a period of no more than four, I will show that the fourth term in the series has to be equal to the seed, or R(4)=seed. To begin with. notice that any number from 0 to 16 can be written as 4n + b where n and b are 0, 1, 2, or 3. Thus:

R(1) = 5*seed mod 16

R(1) = (4 + 1)(4n + b) mod 16

R(1) = (16n +4n +5b) mod 16

Since the 16n is an integer multiple of 16 it contributes nothing to the modulus, and we are left with:

R(1) = (4n +5b) mod 16

While both 4n and 5b are less than 16, their sum may, in general, be more. Thus the modulus can be written as:

R(1) = 4n + 5b - 16a

Where a=1 if 4n + 5b > 16 and is 0 otherwise. We can now find the second term:

R(2) = (4 + 1)(4n + 5b - 16a) mod 16

R(2) = (16n +4n + 20b + 5b - 5*16a) mod 16

R(2) = ( 16n + 16b + 4n + 4b + 5b - 5*16a) mod 16

The first two and the last term are all integer multiples of sixteen, thus their modulus is zero and we are left with:

R(2) = (4n + 4b + 5b) mod 16

Again, although each individual term in this sum is less than sixteen, their sum may not be. Thus to represent the modulus we must include a term that will bring the total to be less than sixteen.

R(2) = 4n + 4b + 5b - 16a

where a is an integer whose value need not be specified because, as has been demonstrated previously, it will fall out when we take the mod 16 in the next step. We are now ready to find the third element in the series and continue as before:

R(3) = (4 + 1)(4n +4b + 5b - 16a) mod 16

R(3) = (16n +16b + 20b - 5*16a + 4n + 4b + 5b) mod 16

R(3) = (16n +16b + 16b - 5*16a + 4n + 4b + 4b + 5b) mod 16

R(3) = (16n + 2*16b - 5*16a + 4n + 8b + 5b) mod 16

Dropping the integral multiples of sixteen as before gives:

R(3) = (4n + 8b + 5b) mod 16

As before the modulus gives us a - 16a term leaving:

R(3) = 4n + 8b + 5b - 16a

Now we use this to find the fourth term:

R(4) = (4 + 1)(4n + 8b + 5b - 16a) mod 16

R(4) = (16n + 32b + 20b - 5*16a + 4n + 8b + 5b) mod 16

R(4) = (16n + 2*16b + 16b - 5*16a + 4n + 8b + 4b + 5b) mod 16

R(4) = (16n + 3*16b - 5*16a + 4n + 17b) mod 16

R(4) = (16n + 4*16b - 5*16a + 4n + b) mod 16

Again we can drop the integral multiples of 16:

R(4) = (4n + b) mod 16

But 4n + b = seed, which is less than 16, so the modulus is simply:

R(4) = 4n + b = seed

Thus the fourth term is equal to the original seed, and the series must begin again. Hence the series repeats with a period of four. Since this was done for any possible seed, we know that all series produced from this algorithm have a maximum period of four.