Number of cases #1

Reading the question, you will realize that the heximal will only contain $1,~2,~3,~4,~5$ .

Let's define a "good" heximal, which is an $n$ -digit heximal that contains only non-zero digits, where every pair of neighboring digits in it has a difference of 1.

$A_n$ is the number of "good" $n$ -digit heximals that end with either $1$ or $5$ .

$B_n$ is the number of "good" $n$ -digit heximals that end with either $2$ or $4$ .

$C_n$ is the number of "good" $n$ -digit heximals that end with $3$ .

$n$ is a non-negative integer.

Then we know that what we're trying to get is $A_{10}+B_{10}+C_{10}$ .

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Then, for convenience, an $n$ -th digit of a number implies it is from the right.

For example, $51354$ 's second digit is $5$ , and fourth digit is $2$ .

Let $k$ be a natural number.

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$i)$ If there's $1$ or $5$ at the $k$ -th digit, the $(k-1)$ -th digit must be $2$ or $4$ , therefore:

$A_k=B_{k-1}$

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$ii)$ If there's $2$ at the $k$ -th digit, the $(k-1)$ -th digit must be $1$ or $3$ , and if there's $4$ at the $k$ -th digit, the $(k-1)$ -th digit must be $3$ or $5$ , therefore:

$B_k=A_{k-1}+2C_{k-1}$

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$iii)$ If there's $3$ at the $k$ -th digit, the $(k-1)$ -th digit must be $2$ or $4$ , therefore:

$C_k=B_{k-1}$

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From $i)$ to $iii)$ , we get:

$B_{k+1}=A_k+2C_k=B_{k-1}+2B_{k-1}=3B_{k-1} \\ \\ B_1=2,~B_2=4$

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Therefore, $B_{2k-1}=2\cdot3^{k-1}$ , and $B_{2k}=4\cdot3^{k-1}$ .

$A_n+B_n+C_n=B_n+2B_{n-1}$

$\therefore~A_{10}+B_{10}+C_{10}=B_{10}+2B_9=4\cdot3^4+2\cdot2\cdot3^4=\boxed{648}$ .