Using the digits 0,1,2,3,& 4, the number of ten digit sequences can be written so that the difference between any two consecutive digits is 1, is equal to `k`, then `k/72` is equal to

Explanation

To solve the problem of finding the number of ten-digit sequences using the digits 0, 1, 2, 3, and 4 such that the difference between any two consecutive digits is 1, we can approach it step by step.

### Step 1: Understand the Constraints
The digits we can use are 0, 1, 2, 3, and 4. The condition that the difference between any two consecutive digits is 1 means that if we are at digit \(d\), the next digit can only be \(d-1\) or \(d+1\).

### Step 2: Define the Problem Recursively
Let \(a_n(d)\) be the number of valid sequences of length \(n\) that end with digit \(d\). The valid digits are 0, 1, 2, 3, and 4. The recursive relations based on the allowed transitions are:
- \(a_n(0) = a_{n-1}(1)\) (can only come from 1)
- \(a_n(1) = a_{n-1}(0) + a_{n-1}(2)\) (can come from 0 or 2)
- \(a_n(2) = a_{n-1}(1) + a_{n-1}(3)\) (can come from 1 or 3)
- \(a_n(3) = a_{n-1}(2) + a_{n-1}(4)\) (can come from 2 or 4)
- \(a_n(4) = a_{n-1}(3)\) (can only come from 3)

### Step 3: Base Case
For \(n = 1\), we have:
- \(a_1(0) = 1\)
- \(a_1(1) = 1\)
- \(a_1(2) = 1\)
- \(a_1(3) = 1\)
- \(a_1(4) = 1\)

So, \(a_1(0) + a_1(1) + a_1(2) + a_1(3) + a_1(4) = 5\).

### Step 4: Compute for \(n = 2\) to \(n = 10\)
Using the recursive relations, we can compute the values for \(n\) from 2 to 10.

- For \(n = 2\):
- \(a_2(0) = a_1(1) = 1\)
- \(a_2(1) = a_1(0) + a_1(2) = 1 + 1 = 2\)
- \(a_2(2) = a_1(1) + a_1(3) = 1 + 1 = 2\)
- \(a_2(3) = a_1(2) + a_1(4) = 1 + 1 = 2\)
- \(a_2(4) = a_1(3) = 1\)

Total for \(n = 2\): \(1 + 2 + 2 + 2 + 1 = 8\).

- Continue this process up to \(n = 10\).

### Step 5: Calculate Total for \(n = 10\)
After calculating recursively up to \(n = 10\), we sum \(a_{10}(0) + a_{10}(1) + a_{10}(2) + a_{10}(3) + a_{10}(4)\) to find \(k\).

### Step 6: Find \(k/72\)
Once \(k\) is calculated, divide it by 72 to find the final answer.