sw00874math-3at4

22


 There are two ways three adjacent integers can be related to "4"
that are of interest to me.  The one way I am not choosing first
begins at zero.  In the 6n Elimination Table the odd squares
always occur in the +1 column, the column beginning with "7".
In today's entry, "4" controls the distribution, but the first n is "1",
and each n is a pown integer which is 2 > the preceding n, and "2"
is the multiplier, and the odd squares always occur in the -1 column,  
 and there is only one row.
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1 2 3   5 6 7   9 10 11   13 14 15   17 18 19   21 22 23   25 26 27

29 30 31   33 34 35   37 38 39   41 42 43   45 46 47   49 50 51

53 54 55   57 58 59   61 62 63   65 66 67   69 70 71   73 74 75

77 78 79   81 82 83   85 86 87   89 90 91   93 94 95   97 98 99

101 102 103   105 106 107   109 110 111   113 114 115   117 118 119

121 122 123   125 126 127   129 130 131   133 134 135   137 138 139

141 142 143   145 146 147   149 150 151   153 154 155   157 158 159

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final-digit rotations:  1 5 9 3 7; 2 6 0 4 8; 3 7 1 5 9

Since this row consists of groups of three adjacent integers,
the first group having a prime pair is:  5 6 7

Beginning with 5 6 7, every group which is greater by "12"
than the group preceding it/ might--except for those groups
in which "4" or "6" is the final digit--contain a prime pair.
The final-digit rotation is: 6 8 0 2 4.              

Here are the first four:  5 6 7   17 18 19   29 30 31   41 42 43

Beginning with 53 54 55, problems:  65 66 67   77 78 79   89 90 91
do not contain prime pairs because of "5" in the first two, and of "7"
in the second two.  In both cases "11" and "13" are the complicit
multiplicands.

Then comes 101 102 103, wherein 101 and 103 is a prime pair.

Then comes 137 138 139, wherein 137 and 139 is a prime pair.

Then comes 149 150 151, wherein 149 and 151 is a prime pair.

Then comes 161 162 163, but/ 7 x 23 = 161.
Notice: 23 - 11 = 12; 25 - 13 = 12.  Okay, backtrack.
5 x 23 = 115   5 x 25 = 125   7 x 23 = 161   7 x 25 = 175

Further:  175 - 70 = 105; and 175 - 14 = 161, and 105 - 14 = 91
Also:  105 + 14 = 119, and 175 + 14 = 189
So these:  3 x 7, 5 x 7, 7 x 7; 13 x 7, 15 x 7, 17 x 7; 23 x 7, 25 x 7, 27 x 7
[ This means that "5" controls distribution--downwards and upwards--
for every pown including itself via a 1 x, 3 x, 5 x, 7 x, 9 x rotation of
single digits and final digits. ]
The node number "105" is a good small example of this/ because it
isdivisible by 3 and 5 and 7, forcing "3" to 102 and 108; forcing "7"
to 98 and 112; and thereby allowing 101, 103, 107, and 109 to be
prime numbers.

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"107" and "109" bring me to the second way, which I will begin
at "4" instead of at "0".  The rotation is: 4 8 2 6 0

3 4 5   7 8 9   11 12 13   15 16 17   19 20 21   23 24 25   27 28 29

31 32 33   35 36 37   39 40 41   43 44 45   47 48 49   51 52 53   55 56 57

59 60 61   63 64 65   67 68 69   71 72 73   75 76 77   79 80 81   83 84 85

87 88 89   91 92 93   95 96 97   99 100 101   103 104 105   107 108 109

111 112 113   115 116 117   119 120 121   123 124 125   127 128 129

131 132 133   135 136 137   139 140 141   143 144 145   147 148 149

151 152 153   155 156 157   159 160 161   163 164 165   167 168 169

In this view the pown squares always occur in the +1 column.

After 3 4 5, wherever a "4" or a "6" is a final digit in a group, the powns
in it cannot be a prime pair.

Beginning with 11 12 13, it appears that only where the pewn of a group
is divisible by "12" can its powns possibly be a prime pair.  In this view
it also appears there are fewer prime pairs.

In the first view the first prime pair is in the 5 6 7 group.  "6" is 6 x 1.
In this view "12" is 6 x 2.  I am deducing from this that in the first
view the pewn in a group containing a prime pair will always be an
odd multiple of "6", whereas in the second view the pewn in a group
containing a prime pair will always be an even multiple of "6". 






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Brian A. J. Salchert