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Old 2013-03-23, 01:45   #1
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Default 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 Euclid-Mullin sequences are generated by the expression Q+1, and that A057207 is generated by 4Q2+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.

4Q2+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. 4Q2-1 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. 4Q2-1 \equiv 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 4Q2-1 factors as (2Q+1)(2Q-1) and that 2Q+1 \equiv 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.
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Old 2013-03-23, 08:16   #2
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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.
Confirmed. The sequence continues 39827899 (Confirmed by TF, though I am sieving the C107 digit cofactor), 11719, 42743, 2143627, 13918999, 169888991, 3343, 1787, 1032463, 38155, 67, 1123.

The next element willl be this P218 unless this 1 (mod 4) C181 should turn out to have 3 (mod) factors.

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though there is still a C283 cofactor which I am currently ECMing.
No result on that after a t35.
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Old 2013-03-23, 17:40   #3
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The next element willl be this P218 unless this 1 (mod 4) C181 should turn out to have 3 (mod) factors.
Should have read 3 (mod 4) factors. It's down to a C142 with still no 3 (mod 4) factor. Stiill a little more ecm to do before it's NFS-ready, which I hereby reserve.

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No result on that after a t35.
Still no result a little under half way to t40.
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Old 2013-03-24, 20:09   #4
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It's down to a C142 with still no 3 (mod 4) factor. Stiill a little more ecm to do before it's NFS-ready, which I hereby reserve.
This turned out not to be necessary because:

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
This was about three-quarters of the way through a t45, so quite lucky. It's also my personal best for ECM, (not that I've done much).
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Old 2013-03-25, 06:38   #5
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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.
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Old 2013-03-25, 12:36   #6
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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.
C251 now fully factored.
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Old 2013-03-26, 21:27   #7
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Confirmed. The sequence continues 39827899 (Confirmed by TF, though I am sieving the C107 digit cofactor), 11719, 42743, 2143627, 13918999, 169888991, 3343, 1787, 1032463, 38155, 67, 1123.
Should read "...11719, 131, 42734..." I am computing these correctly; my mistakes are in transcribing them. The sequence has been approved by oeis, to which you should refer if you want the actual numbers for any reason. Oeis automatically verified the terms on that page against the b-file, which was in turn automatically generated by my scripts, and so isn't prone to this kind of error.

I wanted to see this sequence accepted before proposing the other sequence discussed here, which I'll go ahead with now.
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Old 2013-04-04, 16:52   #8
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I wanted to see this sequence accepted before proposing the other sequence discussed here, which I'll go ahead with now.
Both sequences have now been approved. They are A217759 and A218467

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The next term [of A217759]may be 2336851689743540205539...
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.
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Old 2013-04-11, 12:33   #9
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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.
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Old 2013-04-11, 14:14   #10
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...700@44E6 ... 1017@44E6...
Both of these should of course have been @43E6, not that it makes a hell of a lot of difference. I have been using the recommended ecm limits.
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Old 2013-04-12, 02:44   #11
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Summary of my ecm efforts so far:

Firstly A21847 remains blocked at term 18 after a full t45 + 700@44E6 on the C283 cofactor.
I don't know what gave me that idea. As indicated in the top post, there's a P14 factor congruent to 3 (mod 4). Moreover, both this and

Quote:
...A217759(53) is 4714042483370767...
are easily reached with arbooker's P-1 trick, so are now proven.

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:
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.
Both minus-side and plus-side algebraic factors have proven to be resistant, with just a handful of very small divisors from the former after a t35 on both. Early days.

Last fiddled with by Mr. P-1 on 2013-04-12 at 02:48
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