Within a Diophantine sieve the maximum sieve index is equal to 3G-1 where G is the group number. Then the 6 numbers of each group would have indices: (3G-1,0), (3G-1,1), (3G-2,0), (3G-2,1), (3G-3,0), (3G-3,1). For G=1 they are (2,0), (2,1), (1,1), (1,0), (0,0), (0,1). The sieve index of 0 is (0,0), 2 is (1,0), 3 is (0,1), 4 is (2,0), 5 is (1,1), 7 is (2,1) and the 6 numbers are: 0, 2, 3, 4, 5, 7. For G=20, the sieve indices are (59,0), (59,1), (58,0), (58,1), (57,0), (57,1). These respectively are represented by the 6 numbers: 118, 121, 116, 119, 114, 177, none are known primes. For any given group number G, the two numbers at sieve indices: (3G-3,0) and (3G-3,1), are always divisible by 3. The two at (3G-1,0) and (3G-2,0) are always divisible by 2. Leaving only two prime candidates at sieve indices: (3G-2,1) and (3G-1,1). If both are primes then they are twin primes.
Given a number 12193, its primary sieve index is given by (m, n). For odd numbers, n is always unity, n=1. For even numbers, n is always zero, n=0 while m is any number between 0 and infinity. So, the primary index of 12193 is found by subtracting 3 then dividing by 2 giving 6095 and its primary index is (6095, 1). Next, is to find its group number by assuming that m=6095 is its maximum index value: 3G-1=max(m) then solve for G. The solution gives 2032 and the 6 numbers of this group are: 12186, 12189, 12188, 12191, 12190, and 12193. These are repeated diagonally 2032 times in the Diophantine sieve.


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