mersenneforum.org Sieve for divisibility sequences
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 2017-11-28, 07:44 #1 carpetpool     "Sam" Nov 2016 33510 Posts Sieve for divisibility sequences Prior to other existing sieving programs, it would be convenient to have a sieve for divisiblity sequences S(n) (prime n only) which include, but are not restricted to (a^n-b^n)/(a-b), LucasU(a,b,n), and LucasV(a,b,n). The idea is that there are three arguments to the sieve, S(n) the sequence (already defined), the primes n in the range [a, b], and the sieve limit l, to which S(n) is trial divided by primes of the form 2*k*n+1 (or in some sequences trial divide by primes of the form 2*k*n+-1) with k <= l. The sieve formats should be similar to that of other programs, such as in perl ntheory, msieve, srsieve, etc: [S(n),a,b,l,1] to sieve S(n) for prime indices n, in the range [a,b] and divide by prime numbers of the form 2*k*n+1 up to k = l. [S(n),a,b,l,-1] to sieve S(n) for prime indices n, in the range [a,b] and divide by prime numbers of the form 2*k*n+1 or k*n-1 up to k = l. For instance, if one chooses to test S(n) = 2^n-1 for odd prime n up to 100, with l = 10000, the input is: [2^n-1,3,100,10000,1] and output are the primes: [19, 31, 61, 67, 89] because these primes are not divisible by any prime factor of the form 2*k*n+1 with k < 10000. One can write an ABC file (if there are several primes remaining in output) or test them manually if there are not many and or they are small values: PFGW Version 3.8.3.64BIT.20161203.Win_Dev [GWNUM 28.6] Primality testing 2^19-1 [N-1, Brillhart-Lehmer-Selfridge] 2^19-1 is prime! (0.0667s+0.1339s) Primality testing 2^31-1 [N-1, Brillhart-Lehmer-Selfridge] 2^31-1 is prime! (0.1568s+0.1394s) Primality testing 2^61-1 [N-1, Brillhart-Lehmer-Selfridge] Running N-1 test using base 3 2^61-1 is prime! (0.2173s+0.1997s) Primality testing 2^67-1 [N-1, Brillhart-Lehmer-Selfridge] Running N-1 test using base 3 Primality testing 2^89-1 [N-1, Brillhart-Lehmer-Selfridge] Running N-1 test using base 3 2^89-1 is prime! (0.3988s+0.1553s) and find that 2^n-1 is prime for n = 2, 3, 5, 7, 11, 13, 17, 19, 31, 61, 89. The first 6 terms are a trivial result of dividing 2^n-1 by all prime factors of the form 2*k*n+1, with k > l, and 2*l*n+1 > 2^n-1. The others are a result of a primality test after 'sieving'. Other large sequences should work the same way, leaving less primes indices n to test in S(n).
2022-02-08, 05:20   #2
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

3,643 Posts

Quote:
 Originally Posted by carpetpool and find that 2^n-1 is prime for n = 2, 3, 5, 7, 11, 13, 17, 19, 31, 61, 89. The first 6 terms are a trivial result of dividing 2^n-1 by all prime factors of the form 2*k*n+1, with k > l, and 2*l*n+1 > 2^n-1. The others are a result of a primality test after 'sieving'. Other large sequences should work the same way, leaving less primes indices n to test in S(n).
2^11-1 = 23 * 89 is composite

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