Re: Two large consecutive smooth numbers

Expand Messages
• Jim White & I have been trying to construct these things because they are grist for my new factoring algorithm SMODA. That was one example of our output.
Message 1 of 17 , Mar 4, 2012
Jim White & I have been trying to construct these things because they are grist for my
new factoring algorithm SMODA. That was one example of our output. Although we can break (and have broken) that record, I could only make N have about 30% more digits before
my current program would get very slow or self destruct.

If you have new approaches, and can break that record using them, great.
The PTE approach Broadhurst & Carmody were hinting at, is what I am using now for the
largest ones, so that's not a new idea unless you know something I do not about it.

If you are interested in donating computer time to this effort, intel last 10 years,
unix variant, ok to run weeks at a time in background, then
email warren.wds AT gmail.com.

Thank you.
• ... Yes. Let {f(y)= (y^2-11^4)*(y^2-35^2)*(y^2-47^2)* (y^2-94^2)*(y^2-146^2)*(y^2-148^2)/ 67440294559676054016000 - 1;} With N = f(1210851834572), we find that
Message 2 of 17 , Mar 4, 2012
"WarrenS" <warren.wds@...> wrote:

> N=43623575184339996059537425773119366447006380455838\
> 696504055889999185302903791148393125043181272726633463298672436846034128
> and N+1, both are "smooth", i.e both factor entirely into
> primes <= 9168769.
> Can you do better (i.e. make N larger, and the max prime smaller)?

Yes. Let
{f(y)=
(y^2-11^4)*(y^2-35^2)*(y^2-47^2)*
(y^2-94^2)*(y^2-146^2)*(y^2-148^2)/
67440294559676054016000 - 1;}

With N = f(1210851834572),
we find that N*(N+1) is 5205793-smooth.

With N = f(1606741747790),
we find that N*(N+1) is 8686687-smooth.

David
• ... I have not been following this in detail, but I gained the impression that Warren s original advert was far too optimistic and that now his heuristic for
Message 3 of 17 , Mar 4, 2012
"WarrenS" <warren.wds@...> wrote:

> Jim White & I have been trying to construct these things
> because they are grist for my new factoring algorithm SMODA.

I have not been following this in detail, but I gained the impression
that Warren's original advert was far too optimistic and that now
his heuristic for oracular factorization has escalated from
exp(log(N)^(1/3+o(1))) to the far less encouraging
exp(log(N)^(2/3+o(1))) as the time to construct a database.

Is this a fair summary of the setback?

David
• ... --well, yes and no. (And it wasn t a setback since I knew it all along.) (1) My factorization algorithm SMODA as far as I know still works and still
Message 4 of 17 , Mar 4, 2012
>
>
>
> "WarrenS" <warren.wds@> wrote:
>
> > Jim White & I have been trying to construct these things
> > because they are grist for my new factoring algorithm SMODA.
>
> I have not been following this in detail, but I gained the impression
> that Warren's original advert was far too optimistic and that now
> his heuristic for oracular factorization has escalated from
> exp(log(N)^(1/3+o(1))) to the far less encouraging
> exp(log(N)^(2/3+o(1))) as the time to construct a database.
>
> Is this a fair summary of the setback?

--well, yes and no. (And it wasn't a "setback" since I knew it all along.)

(1) My factorization algorithm SMODA as far as I know still works and still runs in
exp(log(N)^(1/3+-o(1))) time PROVIDED database ("oracle") is available for its use.
It is plausibly better under this proviso than quadratic sieve and number field sieve,
but that at present is unconfirmed.

(2) For the problem of computing the database, however, I only have
exp(log(N)^(2/3+-o(1))) time algorithms for. However as we just saw, the o(1)
is fairly beneficial, since we can reach at least 400-bit-long database entries on a single computer, indeed Broadhurst just found some database entries of that size
in a matter of a few hours -- pretty fast turnaround! (His weren't as large as my
best records, but obviously Broadhurst has already built a search code comparable to or better than mine.) In fact I hope to release a preliminary database by me & Jim White, going up to 400-bits, in a few more days to interested parties.

(3) You might say that (2) sort of demolishes (1), but that is debatable. The thing is,
the database-build is something that all factorers worldwide can do collaboratively and do only once. Therefore, it is not fair to judge this runtime on the same footing as the other runtime. I admit I'm not quite sure how to judge it, because it has been a fairly rare thing
in the world so far, to have oracle-algorithms that actually are useful.

It is conceivable that (2)'s theoretical runtime can be sped up, but at present, I haven't been able to. Furthermore, few or no experts have carefully examined either (1) or (2) yet so it remains possible I'm crazy and the whole thing is broken. I doubt that -- I think any remaining errors are minor -- I'm just giving you fair warning.

--Warren D Smith
• Hard puzzle, really hard puzzle   We know also that, while max N might exist with ~5000 digits, his nearest p-smooth neighbour pair might well be hundreds of
Message 5 of 17 , Mar 5, 2012
Hard puzzle, really hard puzzle

We know also that, while max N might exist with ~5000
digits, his nearest p-smooth neighbour pair might
well be hundreds of digits smaller.

What we don't know is which pairs N are "PTE-compatible",
ie can be found via some factoring
polynomial whose roots are all p-smooth.

Any ideas on that issue would be useful

Jim White

________________________________
From: Andrey Kulsha <Andrey_601@...>
Sent: Sunday, 4 March 2012, 9:53
Subject: Re: [PrimeNumbers] Two large consecutive smooth numbers

Heuristically, log(max_N) is nearly proportional to sqrt(max_prime).

So, with p < 9168769, one can find N with more than 5000 digits.

But that's a hard puzzle, really.

[Non-text portions of this message have been removed]
• Andrey s chain puzzle is interesting.  Could it be he already has found the maximum possible result for chain length 13?   It s hard to see how that result
Message 6 of 17 , Mar 5, 2012
Andrey's chain puzzle is interesting.  Could it be
he already has found the maximum possible result
for chain length 13?

It's hard to see how that result can be beaten.

Some results with weights of 2.2 or more:

28246112570058, weight = 2.2053 (P =  1257251)
18911412089528, weight = 2.2077 (P =  1032307)
218381019281507, weight = 2.2410 (P =  2504167)
9288363679368, weight = 2.2480 (P =   587149)
3393509932556102, weight = 2.2536 (P =  7788997)
4532039198639948, weight = 2.2536 (P =  8856259)
4532039198639949, weight = 2.2536 (P =  8856259)
12469670986534198, weight = 2.2547 (P = 13762769)
10160468895884110, weight = 2.2592 (P = 12163843)
461881571558141, weight = 2.2615 (P =  3050603)
7909529450841510, weight = 2.2621 (P = 10669823)
211814723372355, weight = 2.2918 (P =  1782043)
430753934627814, weight = 2.4217 (P =  1103933)

Perhaps the 14-chain at N = 4532039198639948 might
be a good result? What are the best known results
for 14 or longer chains?

________________________________
From: Andrey Kulsha <Andrey_601@...>
Sent: Sunday, 4 March 2012, 9:53
Subject: Re: [PrimeNumbers] Two large consecutive smooth numbers

> Puzzle: find a chain of 13 consecutive p-smooth integers,
> starting at N, with log(N)/log(p) greater than
>
> log(8559986129664)/log(58393) = 2.71328

Best regards,

Andrey

[Non-text portions of this message have been removed]
• ... No, I think that log/log ratio has no limit. ... Brute force search yielded: N = 505756884840 for 14-chain N = 285377140980 for 15-chain N = 32290958458
Message 7 of 17 , Mar 5, 2012
> Andrey's chain puzzle is interesting. Could it
> be he already has found the maximum possible
> result for chain length 13?

No, I think that log/log ratio has no limit.

> Perhaps the 14-chain at N = 4532039198639948
> might be a good result? What are the best known
> results for 14 or longer chains?

Brute force search yielded:
N = 505756884840 for 14-chain
N = 285377140980 for 15-chain
N = 32290958458 for 16-chain
as listed in http://www.primefan.ru/stuff/math/maxs.xls
(there k+1 is chain length)

Best regards,

Andrey
• Andrey,   I can t use that file, I don t have XL.  Any chance of a text export? eg comma-separated fields     ________________________________ From:
Message 8 of 17 , Mar 5, 2012
Andrey,

I can't use that file, I don't have XL.  Any chance
of a text export? eg comma-separated fields

________________________________
From: Andrey Kulsha <Andrey_601@...>
Sent: Tuesday, 6 March 2012, 6:28
Subject: [PrimeNumbers] Re: 13-chains of consecutive smooth numbers

> Andrey's chain puzzle is interesting. Could it
> be he already has found the maximum possible
> result for chain length 13?

No, I think that log/log ratio has no limit.

> Perhaps the 14-chain at N = 4532039198639948
> might be a good result? What are the best known
> results for 14 or longer chains?

Brute force search yielded:
N = 505756884840 for 14-chain
N = 285377140980 for 15-chain
N = 32290958458 for 16-chain
as listed in http://www.primefan.ru/stuff/math/maxs.xls
(there k+1 is chain length)

Best regards,

Andrey

[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.