The structure of each row

First we study each row.
Fig 5.4 is a part of Fig 5.3.

Figure 5.4  

There are four rows that are colored in red $(z=9)$ , blue $(z=10)$ , green $(z=11)$ and yellow $(z=12)$ respectively. We study these rows one by one.

Figure 5.5  

Figure 5.6 

Fig 5.5 present the case of $z=9$ , and this number $9$ is in the rectangle of gray color. If we start with the last number $14$ and go leftward picking up every other number, then we have $10 \leftarrow 11 \leftarrow 12 \leftarrow 13 \leftarrow 14$ 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 $9 \leftarrow 8 \leftarrow 7 \leftarrow 4 \leftarrow 1$ which starts as an arithmetic sequence with common difference of $3$ and later it becomes an arithmetic sequence with common difference of $1$ .
In this way numbers in Fig 5.5 can be seen as a combination of two different sequences, and we can see this as a decreasing sequence;
$10 \leftarrow 11 \leftarrow 12 \leftarrow 13 \leftarrow 14$
$\swarrow$
$9 \rightarrow 8 \rightarrow 7 \rightarrow 4 \rightarrow 1$ .
Here we present a prediction for the row when $z = n = 1 \ (mod \ 4)$ . Fig 5.6 present the case that is a generalization of the case of $z=n=9$ in Fig 5.5.

Prediction 5.1   Suppose that $z = n = 1 \ (mod \ 4)$ . Then the first number from the right end of the row is $\frac{3n+1}{2}$ . If we start with $\frac{3n+1}{2}$ and go leftward picking up every other number, then we have $n+1 \leftarrow ... \frac{3n-3}{2} \leftarrow \frac{3n-1}{2} \leftarrow \frac{3n+1}{2}$ 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 $\lceil n/4 \rceil$ , where $\lceil \ \ \rceil$ is a ceiling function, and hence the last number of this sequence is $1+3(\lceil n/4 \rceil-1)$ $= 3\lceil n/4 \rceil-2$ . This number $3\lceil n/4 \rceil-2$ is the first number of an arithmetic sequence with common difference of $1$ and the last number of this sequence is $n$ .
In this way numbers in Fig 5.6 can be seen as a combination of two different sequences, and we can see this as a decreasing sequence;
$n+1 \leftarrow n+2 \leftarrow ... \leftarrow \frac{3n-3}{2} \leftarrow \frac{3n-1}{2} \leftarrow \frac{3n+1}{2}$
$\swarrow$
$n \rightarrow n-1 \rightarrow n-2 \rightarrow ... \rightarrow (3\lceil n/4 \rceil-2) \rightarrow ...\rightarrow 7 \rightarrow 4 \rightarrow 1$ .

Figure 5.7  

Figure 5.8  

Similarly Fig 5.7 present the case of $z=10$ , and this number $10$ is in the rectangle of gray color. If we start with the last number $15$ and go leftward picking up every other number, then we have $10 \leftarrow 11 \leftarrow 12 \leftarrow 13 \leftarrow 14 \leftarrow 15$ 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 $9 \leftarrow 8 \leftarrow 7 \leftarrow 4 \leftarrow 1$ , which starts as an arithmetic sequence with common difference of $3$ and later it becomes an arithmetic sequence with common difference of $1$ .
In this way numbers in Fig 5.7 can be seen as a combination of two different sequences, and we can see this as a decreasing sequence;
$10 \leftarrow 11 \leftarrow 12 \leftarrow 13 \leftarrow 14 \leftarrow 15$
$\searrow$
$9 \rightarrow 8 \rightarrow 7 \rightarrow 4 \rightarrow 1$ . Here we present a prediction for the row when $z = n = 2 \ (mod \ 4)$ . Fig 5.8 present the case that is a generalization of the case of $z=n=10$ in Fig 5.7.

Prediction 5.2   Suppose that $z = n = 2 \ (mod \ 4)$ . Then the first number from the right end of the row is $\frac{3n}{2}$ . If we start with $\frac{3n}{2}$ and go leftward picking up every other number, then we have $n \leftarrow ... \frac{3n-4}{2} \leftarrow \frac{3n-2}{2} \leftarrow \frac{3n}{2}$ 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 $\lceil n/4 \rceil$ , and hence the last number of this sequence is $3\lceil n/4 \rceil-2$ . This number $3\lceil n/4 \rceil-2$ is the first number of an arithmetic sequence with common difference of $1$ and the last number of this sequence is $n-1$ .
In this way numbers in Fig 5.8 can be seen as a combination of two different sequences, and we can see this as a decreasing sequence;
$n \leftarrow n+1 \leftarrow ... \leftarrow \frac{3n-4}{2} \leftarrow \frac{3n-2}{2} \leftarrow \frac{3n}{2}$
$\searrow$
$n-1 \rightarrow n-2 \rightarrow n-3 \rightarrow ... \rightarrow (3\lceil n/4 \rceil-2) \rightarrow ...\rightarrow 7 \rightarrow 4 \rightarrow 1$ .

Similarly Fig 5.9 present the case of $z=11$ , and this number $11$ is in the rectangle of gray color. If we start with the last number $17$ and go leftward picking up every other number, then we have $12 \leftarrow 13 \leftarrow 14 \leftarrow 15 \leftarrow 16 \leftarrow 17$ 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 $11 \leftarrow 10 \leftarrow 9 \leftarrow 7 \leftarrow 4 \leftarrow 1$ , which starts as an arithmetic sequence with common difference of $3$ and the last number is 7, and the next number is 9. Note that the difference between $7$ and $9$ is $2$ . After that it becomes an arithmetic sequence with common difference of $1$ .
In this way numbers in Fig 5.9 can be seen as a combination of two different sequences, and we can see this as a decreasing sequence;


$12 \leftarrow 13 \leftarrow 14 \leftarrow 15 \leftarrow 16 \leftarrow 17$
$\swarrow$
$11 \rightarrow 10 \rightarrow 9 \rightarrow 7 \rightarrow 4 \rightarrow 1$ .

Figure 5.9  

Figure 5.10  

Here we present a prediction for the row when $z = n = 3 \ (mod \ 4)$ . Fig 5.10 present the case that is a generalization of the case of $z=n=11$ in Fig 5.9.

Prediction 5.3   Suppose that $z = n = 3 \ (mod \ 4)$ . Then the first number from the right end of the row is $\frac{3n+1}{2}$ . If we start with $\frac{3n+1}{2}$ and go leftward picking up every other number, then we have $n+1 \leftarrow ... \frac{3n-3}{2} \leftarrow \frac{3n-1}{2} \leftarrow \frac{3n+1}{2}$ 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 $\lceil n/4 \rceil$ , and hence the last number of this sequence is $3\lceil n/4 \rceil-2$ . The next number is $3\lceil n/4 \rceil$ that is the first number of an arithmetic sequence with common difference of $1$ and the last number of this sequence is $n$ .
Note that the difference of $3\lceil n/4 \rceil-2$ and $3\lceil n/4 \rceil$ is 2.
Numbers in Fig 5.10 can be seen as a combination of two different sequences, and we can see this as a decreasing sequence;
$n+1 \leftarrow ... \leftarrow \frac{3n-3}{2} \leftarrow \frac{3n-1}{2} \leftarrow \frac{3n+1}{2}$
$\swarrow$
$n \rightarrow n-1 \rightarrow n-2 \rightarrow ...3\lceil n/4 \rceil \rightarrow 3\lceil n/4 \rceil-2 \rightarrow ...\rightarrow 7 \rightarrow 4 \rightarrow 1$ .

Here we present a prediction for the row when $z = n = 0 \ (mod \ 4)$ . Fig 5.12 present the case that is a generalization of the case of $z=n=12$ in Fig 5.11. We omit the explanation on Fig. 5.11, since it is very similar to that of Fig. 5.9.

Prediction 5.4   Suppose that $z = n = 0$ $(mod$ $4)$ . Then the first number from the right end of the row is $\frac{3n}{2}$ . If we start with $\frac{3n}{2}$ and go leftward picking up every other number, then we have $n \leftarrow ... \frac{3n-4}{2} \leftarrow \frac{3n-2}{2} \leftarrow \frac{3n}{2}$ 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 $\lceil n/4 \rceil$ , and hence the last number of this sequence is $3\lceil n/4 \rceil-2$ . The next number is $3\lceil n/4 \rceil$ that is the first number of an arithmetic sequence with common difference of $1$ and the last number of this sequence is $n-1$ .
Note that the difference of $3\lceil n/4 \rceil-2$ and $3\lceil n/4 \rceil$ is 2.
Numbers in Fig 5.10 can be seen as a combination of two different sequences, and we can see this as a decreasing sequence;
$n \leftarrow n+1 \leftarrow ... \leftarrow \frac{3n-4}{2} \leftarrow \frac{3n-2}{2} \leftarrow \frac{3n}{2}$
$\swarrow$
$n-1 \rightarrow n-2 \rightarrow n-3 \rightarrow ...3\lceil n/4 \rceil \rightarrow 3\lceil n/4 \rceil-2 \rightarrow ...\rightarrow 7 \rightarrow 4 \rightarrow 1$ .

Figure 5.11  

Figure 5.12  

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