The Four Survivalists

The Four Survivalists, also known as the List Problem, is a math problem devised by Gilbert Martinez in October 2015. It is still being worked on by Martinez.

In Numbers
How many ways can you write a list containing the numbers 1, 2, 3, and 4 such that the list satisfies the following conditions:
 * At no point does the same number appear consecutively in the list.
 * All numbers are shown in the list the same number of times.

Moreover, how many ways are there to write such a list overall?

In Words
The word version of this problem gives rise to the problem's name.

Zeke, Will, Frank, and Carol are away on a survival trek. Each of them are assigned a certain task every day. The tasks are: cook, scout, guard, and hunter. Here are the following conditions:
 * 1) A period (p) is the number of days that pass before each person is assigned a task a certain number (n) of times.
 * 2) No one can hold the same position for two consecutive days until a period is concluded.

Here are the questions:
 * 1) How many possible configurations can be developed to assign each person one task, given they can be assigned a task n times?
 * 2) How many configurations follow the conditions listed above?

Definitions
Based on the conditions, here's how Martinez defined a period and a configuration.

Period

 * $$p = 4n$$

Let n be the number of times a person is assigned a task.

Configuration

 * $$C_{n} = \frac{p!}{n}\, = \frac{(4n)!}{n}\,$$

Let n = 1

 * $$p = 4n = 4*1 = 4$$


 * $$C_{1} = \frac{4!}{1}\, = 4! = 24$$

Here are all of the configurations for n = 1, which we will call c1.

ZWFC, ZCFW, ZFCW, ZWCF, ZCWF, ZFWC, WZFC, WCFZ, WFCZ, WZCF, WCZF, WFZC, FZCW, GWCZ, FCWZ, FZWC, FWZC, FCZW, CZFW, CWFZ, CFWZ, CZWF, CWZF, and CFZW.

Thus, |c1| = 24.

All 24 configurations follow the conditions.

Let n = 2

 * $$p = 4n = 4*2 = 8$$


 * $$C_{2} = \frac{8!}{2}\, = \frac{40,320}{2}\, = 20,160$$

Using c1, there are 408 configuration permutations that follow the conditions.

For every configuration in c1, there are 17 ways to create a unique permutation. There are 24 configurations in c1, therefore 17 * 24 = 408.

There are 24 additional configurations that can be created with the following 12 items:

ZWZW, ZFZF, ZCZC, WZWZ, WFWF, WCWC, FZFZ, FWFW, FCFC, CZCZ, CWCW, and CFCF.

For each of these items, there are 2 other items that can be affixed to it, creating 2 unique permutations. There are 12 items, therefore 12 * 2 = 24.

Adding these permutations together results in 432 configurations, each of which will be grouped into c2; |c2| = 432.

Let n = 3

 * $$p = 4n = 4*3 = 12$$


 * $$C_{3} = \frac{12!}{3}\, = \frac{479,001,600}{3}\, = 159,667,200$$

As of now, |c3| is unknown.

Lower Estimate
The lower estimate of the value of c3 is found with the rate of change between the values of C2 and C3.
 * $$\Delta(C_{2}-C_{3}) = \frac{C_{3}-C_{2}}{C_{2}}\, = \frac{159,647,040}{20,160}\, = 7,919$$

The rate of change is then multiplied by the value of c2 (432), giving it a proportional change equivalent to the change between C2 and C3. This gives us an estimate of the value of c3.
 * $$|c_{3}| \approx \Delta(C_{2}-C_{3})c_{2} = 7,919*432 = 3,421,008$$

The lower estimate of |c3| is 3,421,008.

Upper Estimate
The upper estimate of |c3| is found by the quotient of c2 and C2.
 * $$\frac{c_{2}}{C_{2}}\, = \frac{432}{20,160}\, = \frac{3}{140}\,$$

The quotient is then multiplied by the value of C3.
 * $$|c_{3}| \approx \frac{3C_{3}}{140}\, = 3,421,440$$

The upper estimate of |c3| is 3,421,440.

3,421,008 ≤ |c3| ≤ 3,421,440.

A more precise estimate of |c3| can be found by the average of these two numbers.
 * $$|c_{3}| \approx \frac{3,421,008 + 3,421,440}{2}\, = 3,421,224$$

Therefore, |c3| ≈ 3,421,224.

Let n = 4

 * $$p = 4n = 4*4 = 16$$


 * $$C_{4} = \frac{16!}{4}\, = \frac{20,922,789,888,000}{4}\, = 5,230,697,472,000$$

As of now, the value of |c4| is unknown.

Lower Estimate
The lower estimate of the value of c4 is found with the rate of change between the values of C3 and C4.
 * $$\Delta(C_{3}-C_{4}) = \frac{C_{4}-C_{3}}{C_{3}}\, = \frac{5,230,537,804,800}{159,667,200}\, = 32,759$$

The rate of change is then multiplied by the lower estimate of value of c3 (3,421,440), giving it a proportional change equivalent to the change between C3 and C4. This gives us an estimate of |c4|.
 * $$|c_{4}| \approx \Delta(C_{3}-C_{4})c_{3} = 32,759*3,421,008 = 112,068,801,072$$

The lower estimate of the value of c3 is 112,068,801,072.

Upper Estimate
The upper estimate of the value c4 is found by the quotient of c2 and C2. The quotient is then multiplied by the value of C4.
 * $$|c_{4}| \approx \frac{3C_{4}}{140}\, = 112,086,374,400$$

The upper estimate of |c4| is 112,086,374,400.

112,068,801,072 ≤ |c4| ≤ 112,086,374,400.

A more precise estimate of |c4| can be found by the average of these two numbers.
 * $$|c_{4}| \approx \frac{112,068,801,072 + 112,086,374,400}{2}\, = 112,077,587,736$$

Therefore, |c4| ≈ 112,077,587,736.

General Formulas
For the total configurations, the general formula is:
 * $$C_{n} = \frac{(xn)!}{n}\,$$

Here, x is the number of unique items in the list, and n is the number of times each item must appear in the list.

As of now, there is no general formula for the exact value of |cn|. However, there is a formula for an estimation of the value of |cn|:
 * $$|c_{n}| \approx \frac{\Delta(C_{n-1}-C_{n})c_{n-1}+\frac{3C_{n}}{140}\,}{2}\,$$

This formula was devised by Martinez on April 30, 2016. Of course, this formula assumes x = 4.

Additional Questions
With this question, Martinez also asks if the values for each of these configurations can be described by a specific function; he also asks what the proportions are between the conditioned configurations and the possible configurations as n gets bigger.

Martinez is currently working on finding the answers to these questions.