DEFINITION: Positive integers $a$ and $b$ are Special if $\gcd(a,b) = |a-b|$.
LEMMA$1$: Positive integers $a$ and $b$ are Special if and only if $a \equiv 0 \bmod{|b-a|}$.
PROOF: Suppose that $a$ and $b$ are Special. Then $|b-a|$ divides $a$. Hence $a \equiv 0 \bmod |a-b|$.
Suppose that $a \equiv 0 \bmod |b-a|$. If $a > b$, then $b = a - |b-a|$. If $a < b$, then $b = a + |b-a|$. In either case $b \equiv 0 \bmod |b-a|$. Since
$b-a = \pm |b-a|$ and $|b-a|$ divides both $a$ and $b$, it follows that $\gcd(a,b) = |b-a|$.
DEFINITION: We say that an increasing sequence of positive integers
$\mathbf s = \{s_0,\; s_1,\; s_2,\; \dots,\; s_n\}$ is Special if, for every $0 \le i < j \le n$, $s_i$ and $s_j$ are special.
An immediate consequence of LEMMA$1$ is
THEOREM$1$: Suppose that the increasing sequence
$\mathbf s = \{s_0,\; s_1,\; s_2,\; \dots,\; s_n\}$
is Special and let
$\sigma = \operatorname{lcm}\left\{ s_i \right\}_{i=0}^n$.
then the sequence
$\mathbf s'=\{\sigma,\; \sigma+s_0,\; \sigma+s_1,\; \sigma+s_2,\; \dots,\; \sigma+s_n\}$
is Special
For example, the sequence $\mathbf s = \{2,3,4\}$ is Special and
$\sigma = \operatorname{lcm}\{ 2,3,4 \} = 12$. Then
$\mathbf s' = \{12,14,15,16\}$ is Special.
$\mathbf s''=\{1680, 1692, 1694, 1695, 1696\}$.
ADDENDUM (4/5/2017)
We will present an example from which the more general case should be clear.
Suppose $A < B < C < D < E$ and we wish to test whether or not
$\mathbf S = \{A,B,C,D,E\}$ is Special.
This will be equivalent to showing that the following pairs are special:
$$\begin{array}{cccc}
(A,B) &(A,C) &(A,D) &(A,E) \\
&(B,C) &(B,D) &(B,E) \\
&&(C,D) &(C,E) \\
&&&(D,E) \\
\end{array}$$
Which is equivalent to showing that
$$\begin{array}{cccc}
B-A \mid A & C-A \mid A & D-A \mid A & E-A \mid A \\
& C-B \mid B & D-B \mid B & E-B \mid B \\
&& D-C \mid C & E-C \mid C \\
&&& E-D \mid D
\end{array}$$
We "encode" this as
$$\begin{array}{lrrrr}
A: & B-A & C-A & D-A & E-A \\
B: & & C-B & D-B & E-B \\
C: & && D-C & E-C \\
D: & &&& E-D \\
E:
\end{array}$$
and call the result a GCD table for the sequence (A,B,C,D,E). For example, the GCD table for the sequence $(1680, 1692, 1694, 1695, 1696)$ is
$$\begin{array}{lrrrr}
1680: & 12 & 14 & 15 & 16 \\
1692: & & 2 & 3 & 4 \\
1694: & && 1 & 2 \\
1695: & &&& 1 \\
1696:
\end{array}$$
Note that
$12,14,15,$ and $16$ divide $1680$,
$2,3$ and $4$ divide $1692$,
$1$ and $2$ divide $1694$, and
$1$ divides $1695$.
That is enough information to prove that the sequence
$(1680, 1692, 1694, 1695, 1696)$ is Special.
Note also that
$(2,3,4) = (14-12, 15-12, 16-12)$,
$(1,2) = (3-2, 4-2)$, and
$1 = 2-1$.
The implied pattern is, in fact, always true.
GCD tables are very useful for generating Special sequences.
Consider, for example the Special sequence $(2,3,4)$. Its GCD table is
$$\begin{array}{lrrrr}
2: & 1 & 2 \\
3: && 1\\
4:
\end{array}$$
let's refer to the table on the right side of the semicolons in a GCD table as the GCD pattern. It shouldn't come as a suprise that other Special sequences will have the same GCD pattern. So, we ask ourselves, "When is the following a GCD table of a Special sequence?"
$$\begin{array}{rrrrr}
n : & 1 & 2 \\
n+1 : && 1\\
n+2 :
\end{array}$$
The answer is, when $2 \mid n$. So $\{(2n, 2n+1, 2n+2): n \in \mathbb Z^+\}$ is a family of special sequences.
Can some of the sequences in this family be extended to a Special sequence containing four elements?
$$\begin{array}{rrrrr}
2n : & \color{red}1 & \color{red}2 & x+2 \\
2n+1 : && \color{red}1 & x+1\\
2n+2 : &&& x \\
2n+2+x:
\end{array}$$
We can ignore the red GCD's because of the way that this table was constructed.
They have already been "tested" and they have passed. So we must find an $x$ such that
$$ x \mid 2n+2, \quad x+1 \mid 2n+1, \quad x+2 \mid 2n$$
If $x=1$, then $1 \mid 2n+2,\; 2 \mid 2n+1$ and $ 3 \mid 2n$. This set of relations cannot be satisfied.
If $x=2$, then $2 \mid 2n+2,\; 3 \mid 2n+1$ and $ 4 \mid 2n$.
The relation $2 \mid 2n+2$ is always true.
$3 \mid 2n+1 \implies 3 \mid n+2 \implies n \equiv 1 \pmod 3$
$4 \mid 2n \implies 2 \mid n \implies n \equiv 0 \pmod 2$.
We find that $n \equiv 4 \pmod 6$.
Letting $x=2$ and $n=6m + 4$, we get
$$\begin{array}{rrrrr}
12m+8 : & \color{red}1 & \color{red}2 & \color{red}4 \\
12m+9 : && \color{red}1 & \color{red}3 \\
12m+10: &&& \color{red}2 \\
12m+12:
\end{array}$$
and we see that, for example, $(8,9,10,12)$ is a Special sequence.
Can we extend this GCD pattern? Let's see.
$$\begin{array}{rrrrr}
12n+8 : &\color{red}1 &\color{red}2 & \color{red}4 & x+4 \\
12n+9 : &&\color{red}1 & \color{red}3 & x+3\\
12n+10: &&&\color{red}2 & x+2\\
12n+12: &&&& x \\
12n + 12 + x:
\end{array}$$
So we need to find a positive integer, $x$ such that
$$ x+4 \mid 12n+8,
\quad x+3 \mid 12n+9,
\quad x+2 \mid 12n+10,
\quad x \mid 12n+12
$$
The first good value of $x$ is $12$. Letting $x=12$, we get the relations
$$ 16 \mid 12n+8,
\quad 15 \mid 12n+9,
\quad 14 \mid 12n+10,
\quad 12 \mid 12n+12
$$
-
$12 \mid 12n+12$ is true for all $n$.
-
$ 14 \mid 12n+10
\implies 7 \mid 6n + 5
\implies 7 \mid n+2
\implies n \equiv 5 \pmod 7$
-
$15 \mid 12n+9
\implies 5 \mid 4n+3
\implies 5 \mid n+2
\implies n \equiv 3 \pmod 5$
-
$16 \mid 12n+8
\implies 4 \mid 3n + 2
\implies 4 \mid n + 2
\implies n \equiv 2 \pmod 4$
Using the Chinese remainder theorem, we find $n \equiv 138 \pmod{140}$.
Letting $x=12$ and $n=138m + 140$, we get
$$\begin{array}{rrrrr}
1680m + 1664: & \color{red}1 & \color{red}2 & \color{red}4 & \color{red}{16}\\
1680m + 1665: && \color{red}1 & \color{red}3 & \color{red}{15}\\
1680m + 1666: &&& \color{red}2 & \color{red}{14}\\
1680m + 1668: &&&& \color{red}{12}\\
1680m + 1676:
\end{array}$$
So, we see that $(1664, 1665, 1666, 1668, 1676)$ is a Special sequence.
Addendum 5/29/2017
For the sake of expedience, I offer the following theorems without proof. First, I want to remark that there is no reason a Special sequence can't start with $0$.
\begin{array}{c|ccccc}
x_1 & \delta_{12} & \delta_{13} & \delta_{14} & \cdots & \delta_{1n}\\\hline
x_2 && \delta_{23} & \delta_{24} & \cdots & \delta_{2n}\\\hline
x_3 &&& \delta_{34} & \cdots & \delta_{3n}\\\hline
\vdots
&&&& \ddots & \vdots\\\hline
x_{n-1} &&&&&\delta_{n-1,n} \\\hline
x_n
\end{array}
THEOREM Let $x_1, x_2, x_3, \dots, x_n$ be a strictly increasing sequence of integers. Define $\delta_{ij} = x_j - x_i$. Then $x_1, x_2, x_3, \dots, x_n$ is a Special sequence if and only if, for every $1 \le i < j \le n, \; \delta_{ij} \mid x_i$.
DEFINITION We define the period of a Special sequence to be the number $N$ where $N$ equals the least common multiple of all of the $\delta_{ij}$.
THEOREM Let $(a_1, a_2, \dots, a_n)$ be a Special sequence with a period of $N$. Then $(N+a_1, N+a_2, \dots, N+a_n)$ is also a Special sequence with a period of $N$.
THEOREM Let $(a_1, a_2, \dots, a_n)$ be a Special sequence with a period of $N$ and with $0 < a_1 < a_2 \cdots < a_n$. Then $(0, a_1, a_2, \dots, a_n)$ is also a Special sequence.
THEOREM Let $(a_1, a_2, \dots, a_n)$ be a Special sequence with period $N$. Then $(N-a_n, N-a_{n-1}, \dots, N-a_2, N-a_1)$ is also a Special sequence with a period of $N$. We call this sequence the reflection of the sequence $(a_1, a_2, \dots, a_n)$.
EXAMPLES
$$\begin{array}{c|ccccc}
0 & 1 & 2 \\\hline
1 && 1\\\hline
2
\end{array}$$
We see from the table above that $(0,1,2)$ is a special sequence with a period of $2$ that is equal to its reflection.
Also $2+(0,1,2) = (2,3,4)$ is a Special sequence with a period of $2$.
$$\begin{array}{c|ccccc}
2 & 1 & 2 \\\hline
3 && 1\\\hline
4
\end{array}$$
Hence $(0,2,3,4)$ is a special sequence with a period of $12$.
$$\begin{array}{c|ccccc}
0 & 2 & 3 & 4 \\\hline
2 && 1 & 2\\\hline
3 &&& 1 \\\hline
4
\end{array}$$
Its reflection also has a period of $12$.
$$\begin{array}{c|ccccc}
8 & 1 & 2 & 4 \\\hline
9 && 1 & 3\\\hline
10 &&& 2 \\\hline
12
\end{array}$$
You should have the idea now how to generate ever larger Special sequences.
