4n

4n+1

4n+2

4n+3


9n

36

9

18

27

9n+1

28

1

10

19

9n+2

20

29

2

11

9n+3

12

21

30

3

9n+4

4

13

22

31

9n+5

32

5

14

23

9n+6

24

33

6

15

9n+7

16

25

34

7

9n+8

8

17

26

35

Now I am going to duplicate the 9n +1 and 9n+7 plane. The 9n+2 and 9n+8 plane are coloured black (as well as the 4n+2 column). The 9n+5 is coloured yellow.
4n

4n+1

4n+2

4n+3


9n

36

9

18

27

9n+1

28

1

10

19

9n+1

28

1

10

19

9n+2

20

29

2

11

9n+3

12

21

30

3

9n+4

4

13

22

31

9n+5

32

5

14

23

9n+6

24

33

6

15

9n+7

16

25

34

7

9n+7

16

25

34

7

9n+8

8

17

26

35

If we omit the black planes at 9n+2 and 9n+8 we have two horizontal black planes as in the original diagram (in the proof I talk about merging planes but I think it makes more sense to simply discard them.) The grid contains some repetitions and some numbers in the number chain are missing (but we know they aren't twin primes).
Next we need to extend this up to the 25n plane. The first few slices look like this (before they are sorted by their residue mod 25):
To make it easier to see what the next stage looks like, I'm going to leave out the black planes and the red plane (in which there are no primes other than 3, which is an exception  it isn't a twin because it is only on one red plane. )
So we start with this pattern:
9n

36

9

18

27

9n+1

28

1

10

19

9n+1

28

1

10

19

9n+3

12

21

30

3

9n+4

4

13

22

31

9n+5

32

5

14

23

9n+6

24

33

6

15

9n+7

16

25

34

7

9n+7

16

25

34

7

Next we need to extend this up to the 25n plane. The first few slices look like this (before they are sorted by their residue mod 25):
36

9

18

27

28

1

10

19

28

1

10

19

12

21

30

3

4

13

22

31

32

5

14

23

24

33

6

15

16

25

34

7

16

25

34

7

72

45

54

63

64

37

46

55

64

37

46

55

48

57

66

39

40

49

58

67

68

41

50

59

60

69

42

51

52

61

70

43

52

61

70

43

108

81

90

99

100

73

82

91

100

73

82

91

84

93

102

75

76

85

94

103

104

77

86

95

96

105

78

87

88

97

106

79

88

97

106

79

To make it easier to see what the next stage looks like, I'm going to leave out the black planes and the red plane (in which there are no primes other than 3, which is an exception  it isn't a twin because it is only on one red plane. )
So we start with this pattern:
4n+1

4n+3


9n+1

1

19

9n+1

1

19

9n+4

13

31

9n+4

13

31

9n+5

5

23

9n+7

25

7

9n+7

25

7

Here, the numbers in the green and yellow planes up to 900 have been rearranged by their residue mod 25 (picture this as 25 slices of a cuboid):
Next, we are going to substitute:
25n+3 for 25n+2
25n+13 for 25n+12
25n+18 for 25n+17
25n+ 23 for 25n+22
The first prime that is a nontwin is 11  this has already been removed from the pattern when we coloured the 9n+2 plane black  we chose to repeat the 9n+1 plane, so you could see the second instance of 19 as a replacement for 11.
Every number that is only in one red plane when this process is reiterated in infinite dimensions will also be on only one yellow plane, so we can treat the yellow plane as being the same as a green plane.*
Given this, the intersection of green, black and red planes in the grid above is the same topological pattern as we would find if we were looking only for primes, but we have multiple repetitions of individual numbers within the pattern.
Also, I talk about rearranging the numbers according to modulo residue above  it would maybe be better to visualise this as a hypercuboid that from the start contains duplications of some planes that aren't either pn or pn+2 and doesn't include any pn+2 planes other than the (p^2)n+p+2 planes.
Hope that all makes sense?
In theory we need to make these plane substitutions in every dimension  I do have one or two thoughts on why that might be problematic, I'll come back to talk about that soon.
*See the algebra in part 7 of the original proof.
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