What is the minimum of the largest number in an array such that existing four different indices $a,b,c,d$ which satisfy $A_a+A_b=A_c+A_d$? For all non-descending positive integer arrays $A$ satisfying:

*

*There exist four different indices $a,b,c,d (a \lt c \lt d\lt b)$such that $A_a+A_b=A_c+A_d$ ;

*For all those $a,b,c,d$ s, $|a-c|\ge k$ and $|b-d|\ge k$ ($k\ge1$).

What is the minimum of the largest number in those arrays?
For example, if $k=4$, we can find an array $B = \{1,4,6,7,7,12\}$. The only four different indices $a=0, b=5,c=2,d=3$ simultaneously satisfy $1+12=6+7$ , $|2-0|\ge 2$ and $|5-3|\ge 2$. The largest number in $B$ is $12$. Other arrays like $\{1,5,8,10,11,17\}$ also satisfies but its largest number $17$ is larger than $12$.
However, the array $C = \{1,1,1,1,1,1\}$ is unable to meet the conditions since their exist four indices $a=0, b=3,c=1,d=2$ which satisfy $1+1=1+1$ but not the second condition.
I wonder if $12$ is the minimum of the largest number in those arrays when $k=4$, and what is the minimum when a specific $k$ is given.
 A: Let $\ F(0):=F(1):=1,\ $ and
$\ F(n)=F(n-1)+F(n-2)\ $ for every $\ n=2\ 3\ \ldots.\ $
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Theorem 1
$$ F(a)+F(d)\ > F(b)+F(c) $$
for arbitrary integers $\ a>b>c>d\ge 0.$
This holds even if additionally $\ F(-1):=1\ $ and
$\ d\ge -1$.
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Theorem 2  For $\ n>2\ $ and integers
$$ G(n)\ge G(n-1)\ge\ldots
           \ge G(0)\ge G(-1)\ge 1 $$
such that
$$ G(a)+G(d)>G(b)+G(c) $$
for arbitrary $\ a>b>c>d\ge -1\ $ we have
$$ \forall_{k=0}^n\quad G(k)\ge F(k) $$
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Proof  We may assume that $G(-1)=1$ or
else consider $H(-1):=1\ $ and
$\ H(k):=G(k)+1-G(0)\ $ for every $\ k=0\ldots n$.
Thus let $G(-1):= 1.\ $ Then we may also assume
$G(0):=1$ or
else consider $H(-1):=H(0):=1\ $ and
$\ H(k):=G(k)+1-G(1)\ $ for every
$\ k=1\ldots n$.
Thus let $G(-1):=G(0):=1.\ $ Then we may also
assume $G(1):=1$ or
else consider $H(-1):=H(0):=H(1):=1\ $ and
$\ H(k):=G(k)+1-G(2)\ $ for every
$\ k=2\ldots n$.
Thus assume $G(-1):=G(0):=G(1):=1.$
We have the first two induction steps $\ k=0\ $
and $\ k=1.$
Assume (induction) that
$\ \forall_{i=0}^k\ G(i)\ge F(i)\ $
for a certain $\ k\ge 1.$
Actually, we can reduce this to
$\ \forall_{i=0}^k\ G(i)=F(i).$
Indeed, let $j$ (if any) be the smallest such
that $\ G(j)>F(j).\ $ Then define
$\ H(i):=G(i)\ $ for $\ i=-1\ldots j-1\ $ and
$\ H(i)=G(i)+F(j)-G(j)\ $ for $\ i=j\ldots n$
and the rest is clear. Great!
Thus, let $\ G(i)=F(i)\ $ for $\ i=0\ldots k.\ $
We want to prove the induction step
$\ G(k+1)\ge F(k+1).$
Indeed, 
$$ G(k+1)+1=G(k+1)+G(-1) > G(k)+G(k-1) =
          F(k)+F(k-1) = F(k+1) $$
End of Proof
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Theorem 3  Let $\ A(m)\ $ be the smallest integer such that $\ A(m)=G(n+1)\ $
for arbitrary integers
$$ G(n+1)\ge G(n)\ge\ldots\ge G(-1)\ge 1\ $$
such that $\ n>m\ $ and
$$ G(a)+G(d)\ >\ G(b)+G(c) $$
for every $\ n+1>a>b>c>d\ge-1$ except for
$$ (a\ b\ c\ d)\ =
   \ (n\!+\!1\ \ n\,\ c\,\ d) $$
such that
$\ \{c\ d\}\subset\{1\ \ 0\,\ -\!\!1\}.$
Then   $A(m)\ =\ F(m).$
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Proof
We have more, namely, $\ n=m\ $ and the critical
sequence $\ G\ $ is given as follows:
$\ G(-1):=-1\ $ and
$$ \forall_{k=0}^n\quad G(k):=F(k) $$
and $\ G(n+1):= G(n) = F(n) $
This answers the OP's question (but for routine extra connection with the literal OP's statement).  End of Proof
