- --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
>

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

>

>

> --- In primenumbers@yahoogroups.com,

> "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?

(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 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@...>

To: PrimeNumbers@...

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 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@...>

To: PrimeNumbers@...

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,

Best regards,

> starting at N, with log(N)/log(p) greater than

>

> log(8559986129664)/log(58393) = 2.71328

Andrey

[Non-text portions of this message have been removed] > Andrey's chain puzzle is interesting. Could it

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

> be he already has found the maximum possible

> result for chain length 13?

> Perhaps the 14-chain at N = 4532039198639948

Brute force search yielded:

> might be a good result? What are the best known

> results for 14 or longer chains?

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: Andrey Kulsha <Andrey_601@...>

To: PrimeNumbers@...

Sent: Tuesday, 6 March 2012, 6:28

Subject: [PrimeNumbers] Re: 13-chains of consecutive smooth numbers

> Andrey's chain puzzle is interesting. Could it

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

> be he already has found the maximum possible

> result for chain length 13?

> Perhaps the 14-chain at N = 4532039198639948

Brute force search yielded:

> might be a good result? What are the best known

> results for 14 or longer chains?

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]