Case 2

In this subsection we study the case that $z = 2 (mod \ 4)$ and $y$ is odd.
An example of this case is Example 5.4.
The argument here looks complicated, but it is a generalization of argument used in Example 5.4
.

Figure 5.20  

In Fig. 5.20 we suppose that Predictions 5.1, 5.2, 5.3,5.4 and 5.5 are valid for $z=0,1,2,...,4m+1$ and for $y = 0,1,2,...,z$ . We also suppose that these predictions are valid for $z=4m+2$ and $y=0,1,...,2k-4$ . Under these assumptions we prove that $X= G1R(\{2k-3,4m+2\}) = 4m-k+3$ .

For $z=n=4m+1 = 1 (mod \ 4)$ by Prediction 5.1 the first number from the right end of the row is $6m+2$ . If we start with $6m+2$ and go leftward picking up every other number, then we have $4m+2 \leftarrow ... 6m\leftarrow 6m+1 \leftarrow 6m+2$ which is an arithmetic sequence with common difference of $-1$ .
On the other hand if we start with $1$ that is the second number from the right end of the row and go leftward picking up every other number, then we have an arithmetic sequence with common difference of $3$ whose length is $\lceil (4m+1)/4 \rceil$ $=m+1$ , and hence the last number of this sequence is $3\lceil n/4 \rceil-2$ $=3(m+1)-2 = 3m+1$ . This number $3m+1$ is the first number of an arithmetic sequence with common difference of $1$ and the last number of this sequence is $4m+1$ .
By Prediction 5.5 for $y = n=2k-3$ the number at the top of the column is $\frac{3n+1}{2}$ $=3k-4$ . If we move down picking up every other number, then we have an arithmetic sequence with common difference of $2$ .
On the other hand if we start with 1 that is the second number of the column and move down picking up every other number, then we have $1,4,7,10,13,...$ that starts as an arithmetic sequence with common difference of $3$ such that it has $\lfloor \frac{n}{2} \rfloor +1$ $=k-1$ terms and the last number is $3(k-2) +1=3k-5$ , then it becomes an arithmetic sequence with common difference of $2$ whose first number is $3k-5$ .
In Fig. 5.20 we suppose that

$$G1R(\{2m,4m+1\}) = 3m+1,$$ (5.42)

$$G1R(\{2k-3,4k-6\}) = 3k-5,$$ (5.43)

$$4m+1 > 4k-6 2k-3 < 2m.$$ (5.44)

By Lemma 5.1 we have only to prove that $4m-k+3$ is the smallest number that does not belong to $L1(\{2k-3,4m+2\})$ to get $X= G1R(\{2k-3,4m+2\}) = 4m-k+3$ .
By Definition 5.2 $LG1R(\{2k-3,4m+2\})$ can be separated into three parts $L11(\{2k-3,4m+2\})$ , $L12(\{2k-3,4m+2\})$ and $L13(\{2k-3,4m+2\})$ .
$L11(\{2k-3,4m+2\}) = \{G1R(\{0,0\}),G1R(\{1,1\}),...,G1R(\{2k-3,2k-3\})\}$

$$= \{0,2,3,5,...,3k-6,3k-4\}.$$ (5.45)

$L12(\{2k-3,4m+2\}) = \{G1R(\{2k-3,2k-2\}),G1R(\{2k-3,2k-1\}),...,G1R(\{2k-3,4m+1\})\}$
$=\{1,3k-2,4,...,4m-k+1,4m+k\}$

$$\cup \{1,4,...3k-8,3k-5,3k-3\}$$ (5.46)

$$\cup \{3k-1,3k+1,...,4m-k+1\}$$ (5.47)

$$\cup \{3k-2,3k,...,4m+k\}.$$ (5.48)

$L13(\{2k-3,4m+2\}) = \{G1R(\{0,4m+2\}),G1R(\{1,4m+2\}),...,G1R(\{2k-4,4m+2\})\}$
$=\{4m+2,4m+1,4m+3,...,4m-k+4,4m+k\}$

$$=\{4m+2,4m+3,4m+4,...,4m+k\}$$ (5.49)

$$\cup \{4m+1,4m,4m-1,...,4m-k+4\}.$$ (5.50)

By (5.45) and (5.46) we have $L11(\{2k-3,4m+2\}) \cup L12(\{2k-3,4m+2\}) \supset$ $\{0,1,2,3,...,3k-3\}$ .
By (5.49) and (5.50) all the numbers in $L13(\{2k-3,4m+2\})$ are bigger than $4m-k+3$ .
By (5.47) and (5.48) $L12(\{2k-3,4m+2\})$ $\supset \{3k-2,...,4m-k+2\}$ . We know that the list $\{3k-2,...,4m+k\}$ does not contain $4m-k+3$ , since the numbers in the list and the number $4m-k+3$ have opposite parity. $($ In other words if $4m-k+3$ is odd, then the numbers in the list are even, and if $4m-k+3$ is even, then the numbers in the list are odd. $)$ . Therefore we have $X=4m-k+3$ .