View Single Post
Mar3-12, 03:22 PM
P: 894
As Dodo noted, my prior test excluded some primes of the form 8n +/- 3, e.g. 29. I discovered a new property of the recursive series used in the prior test to allow a correction that apparently includes all primes of the form 8n +/- 3 but apparently includes no composites. The property of the recursive series (defined by S_0 = 2, S_1 = 3, S_n = 6S_(n-1) - S_(n-2) - 6) is that;

[tex]\sum_{i=0}^{2n} \binom{2n}{i}S_{k-n+i} + 3*(2^{n}-1)*2^{2n-1} = 2^{3n} * S_{k}[/tex]

Now if 2n = P+1 and P is prime of the form 8n +/- 3 and k = (P-1)/2, the side on the right = 0 mod P and only the terms [tex]S_{-1}, (P+1)*S_{0},(P+1)*S_{P-1} \{and} S_{p}[/tex] not divisible by P on the left are . My experience is that S(-1) = 3, S(0) = 2, S(P-1) = P*S + 10 and S(P) = P*R + 3. Thus my new conjecture is that If and only If 6*2^((P+1)/2) equals -12 mod P, where P is a number of the form 8n +/- 3, then P is prime. I checked all 45 of my false hits, F, under 3 million, i.e. where S_((F-1)/2) = 0 mod F. However, I hadn't yet gone back and checked whether any composite meets this latter test and also does not have a zero at S_((F-1)/2).
Edit I found ten false hits under 500,000. All were composites F such that S((F-1)/2) > 0 mod F. Thus there is a need to further check that in the series S_0 = 2, S_1 = 3, S_n = 6*S_(n-1) - S_(n-2) - 6 that S((P-1)/2) = 0 mod P to determine that P is prime. None the less, both of these checks take far less time than checking that each and every term S_i is not equal 0 for i < (p-1)/2 plus you have the advantage that all primes of the form 8n +/- 3 apparently meet the test.
Phys.Org News Partner Science news on
Fungus deadly to AIDS patients found to grow on trees
Canola genome sequence reveals evolutionary 'love triangle'
Scientists uncover clues to role of magnetism in iron-based superconductors