20130323, 01:45  #1 
Jun 2003
7·167 Posts 
Two new sequences
Denote by Q the product of the elements so far and let E(Q) be an expression in Q. Say that a sequence is generated by E(Q) if each element is a prime factor of E(Q). Using this terminology we can say, for example, that the EuclidMullin sequences are generated by the expression Q+1, and that A057207 is generated by 4Q^{2}+1. The generator does not completely determine the sequence  it is also necessary to specify the criterion which determines which prime factor out of several becomes the next element. It is not, however, necessary to specify the first element  the generator can be used to calculate the first element using the fact that the empty product is 1.
4Q^{2}+1 has the interesting property that all the sequences it generates, irrespective of the selection criterion, consist solely of primes congruent to 1 (mod 4). This fact follows from quadratic reciprocity. 4Q^{2}1 has no similar virtue. If we choose the smallest prime factor each time, the sequence we get is essentially A102926 with the initial 2 omitted. In this case primes congruent to 1 and to 3 (mod 4) appear seemingly at random. This defect is easily remedied. 4Q^{2}1 3 (mod 4), therefore one or more of its prime factors must also be so congruent. If we use a selection criterion that only selects from these, then the result is an infinite sequences of primes, all congruent to 3 (mod 4). For example if we choose the least such prime, then the sequence we get is 3, 7, 43, 19, 6863, 883, 23, 191, 2927, 205677423255820459, 11, 163, 227, 9127, 59, 31, 71, 131627, 2101324929412613521964366263134760336303, 127, 1302443, 4065403, 107, 2591, 21487, 223, 12823, 167, 53720906651. The next element is probably 5452254637117019, though proof of that depends upon the factorisation of a C103 cofactor, which I am sieving even as I write. Alternatively, noting that 4Q^{2}1 factors as (2Q+1)(2Q1) and that 2Q+1 3 (mod 4), we can choose our 3 (mod 4) factor from that side only, and dispense with the other side altogether. Again, choosing the least such prime gives us the sequence 3, 7, 43, 139, 50207, 23, 10651, 563, 11, 19, 363303615453958067659, 787, 2803, 3261639461817858097484047657974700766171, 448513341328399688966874038187266281752082128599801650127, 89724193529143, P149. The next element is probably 34901491624723, though there is still a C283 cofactor which I am currently ECMing. Neither sequence is in OEIS, and a Google search on some of the larger primes turns up no hits. 
20130323, 08:16  #2  
Jun 2003
7×167 Posts 
Quote:
The next element willl be this P218 unless this 1 (mod 4) C181 should turn out to have 3 (mod) factors. Quote:


20130323, 17:40  #3  
Jun 2003
7×167 Posts 
Quote:
Quote:


20130324, 20:09  #4  
Jun 2003
491_{16} Posts 
Quote:
Code:
Using B1=11000000, B2=35133391030, polynomial Dickson(12), sigma=3562441688 Step 1 took 25294ms Step 2 took 8744ms ********** Factor found in step 2: 28054051051660897912550965912627155892579390168269491 Found probable prime factor of 53 digits: 28054051051660897912550965912627155892579390168269491 Probable prime cofactor 78515723547747536673177045025872562456609996762265415888431059177491610095123108613576699 has 89 digits 

20130325, 06:38  #5 
Jun 2003
2221_{8} Posts 
With other things going on, I failed to follow up on that last post, which is that I can continue the first of my two sequences, provisionally numbered A217759, specifically 28054051051660897912550965912627155892579390168269491, 463, 2459, 6899, 47, 587, 678563, 458191. The next term may be 2336851689743540205539, but there are still C251 and C289 cofactors.
Still no luck on the C283 needed to extend my second sequence, (which I don't yet have an A number for), despite more than a t45. On to t50. 
20130325, 12:36  #6  
Jun 2003
7·167 Posts 
Quote:


20130326, 21:27  #7  
Jun 2003
1169_{10} Posts 
Quote:
I wanted to see this sequence accepted before proposing the other sequence discussed here, which I'll go ahead with now. 

20130404, 16:52  #8  
Jun 2003
7·167 Posts 
Quote:
With one of the algebraic factors fully factored and the other at >t45 I'm prepared to proceed on the assumption that it is. On that basis, the next terms may be 4714042483370767, 13751. I haven't done enough testing on the cofactors to proceed further. About a tenth of the way into a t50, A21847 remains blocked. 

20130411, 12:33  #9 
Jun 2003
7×167 Posts 
Summary of my ecm efforts so far:
Firstly A21847 remains blocked at term 18 after a full t45 + 700@44E6 on the C283 cofactor. The C289 required to prove that A217759(52) is 2336851689743540205539 has had a full t45 plus a further 1017@44E6. Assuming that it is, the two cofactors needed to prove that A217759(53) is 4714042483370767 are a C272 and a C247. Both have had a full t45. Assuming that it is, a C293 is all that stands in the way of proof that A217759(55) is 60594336654053539135723. A full t45 will be completed shortly. All of the above are released. Assuming that all the conjectured terms of A217759 are correct, the next terms will be. 563, 3671, 719, 34607, 60659. I haven't even began to tackle the factorisation for the next term, which I hereby reserve. 
20130411, 14:14  #10 
Jun 2003
7·167 Posts 

20130412, 02:44  #11  
Jun 2003
7×167 Posts 
Quote:
Quote:
The next term of A218467 might be 260121991330279595838004905951666464027, (P39) but as I about a tenth of the way into a t40 when I found it, I'm certainly not confident that there is nothing smaller hidden in the C276 cofactor. Quote:
Last fiddled with by Mr. P1 on 20130412 at 02:48 

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