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ramsey2879
#1
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.
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