- David Broadhurst wrote January 23:
> Thanks for the history, Jens.And the amazing twin record by Daniel Papp is 33218925*2^169690+/-1 with 51090
> > Now I don't have to worry about proofs,
> > I will search for a larger case of 3 consecutive semi-primes.
> Combining pairs of 5k-digit primes in x
> a double-sieved attack on (x+2)/2 and (x+3)/3
> at 10k digits looks feasible. After all,
> it's only a time=O(digits^4) problem,
> and there are already nearly 20 pairs of
> gigantic twins.
digits. This alone seems equivalent to 5.1^4=677 10k-digit solutions. The
multiplier is small and I only see 7 top 5000 primes with that exponent, so I
guess he got very lucky.
However, I enjoy to use my single PC for a lot of varying computational
problems. I rarely let anything run for more than 2 weeks and I want a good
success chance - I know this is not the way to fame and glory but I just like
the satisfaction of reaching my target size, as long as it is not trivial.
A 10k-digit semi-prime triplet looked like an expected 6 months with the
generic prp in PrimeForm and the likely requirement for thousands of 5k-digit
primes on a suitable primorial form. I settled for a modest 4k digits and just
succeeded after a little bad luck.
p=672066*4691#+1 and q=2329500*4691#+1 are primes.
p * q, 2 * (pq+1)/2, 3 * (pq+2)/3 are 4021-digit semi-primes.
I trial factored with my own program. All prp's and proofs were found with
As David noted, the form gives 50% N-1 factorization of (pq+1)/2 and (pq+2)/3
Hmm, I just noticed the prime triplet record is 4135 digits. Not exactly the
same complexity with the right algorithm, but the target size should have been
set to "beat" that. I will try to "beat" the twin record instead. I can
combine primes from an old top 5000 for that. The heuristics for this problem
do not require a primorial factor if there are hundreds of primes available.
Jens Kruse Andersen
- Congrats, Jens, on your semi-primes. My taste is much as
as yours: big enough to be a challenge, small enough to give
a result in O(week). Incidentally the real breakthrough for
> the prime triplet record is 4135 digitswas avoiding ECPP. But this was still an O(digits^5) problem,
whereas yours was O(digits^4). I guess that your slowdown
was that your trial factoring was merely the -f option of Pfgw,
whereas I handcrafted a sieve for the BLS-provable triplets.
- PS: I re-read saw that you did your own trial factoring, sorry.
But I guess that it was not truly a sieve, but just for
one pair at a time? Mind you it's not easy to see a way
round that. If anyone could sieve this two-parameter
input problem, it would be Phil.
- --- "David Broadhurst <d.broadhurst@...>" <d.broadhurst@...> wrote:
> Congrats, Jens, on your semi-primes....
> I guess that your slowdownIn the world of sieving, Jens is one of the people to whom I doff my hat.
> was that your trial factoring was merely the -f option of Pfgw,
> whereas I handcrafted a sieve for the BLS-provable triplets.
He stood out on alt.math.recreational because of his sieving ability.
Is that free as in Willy or as in bird?
Do you Yahoo!?
Yahoo! Mail Plus - Powerful. Affordable. Sign up now.
- About prime and semi-prime tuples I wrote:
> Not exactly the same complexity with the right algorithmDavid Broadhurst wrote:
> Incidentally the real breakthrough for the prime triplet record is 4135"Not exactly" was a deliberate understatement. I know there is a huge
> digits was avoiding ECPP. But this was still an O(digits^5) problem,
> whereas yours was O(digits^4).
difference and it makes little sense to compare sizes, except for fun. The
"wrong" algorithm for k consecutive numbers with the same number of prime
factors (used for a 7-tuple with 3 prime factors by Phil, who admitted to
being sloppy) is: Search k simultaneous prime cofactors, e.g. x/5, (x+1)/2,
(x+2)/3. This was also the suggestion of others in alt.math.recreational and
would give the normal prime tuple complexity O(k+2).
David Broadhurst wrote:
> PS: I re-read saw that you did your own trial factoring, sorry.Yes, I trial factored one pair at a time. My candidates were on the form
> But I guess that it was not truly a sieve, but just for
> one pair at a time? Mind you it's not easy to see a way
> round that. If anyone could sieve this two-parameter
> input problem, it would be Phil.
p*q = (c*6*4691#+1) * (d*6*4691#+1)
This (including the 6) prevents factors below 4691 in both (pq+1)/2 and
I precomputed an array of 6*4691# mod (primes<10^8).
Then I could quickly trial factor each pair from 4691 to 10^8 (a high limit
for individual trial on 4k-digits with a 2.7s prp test) using only 32-bit
inline assembler instructions on c and d.
I thought about sieve possibilities but it seemed hard and with a few days
expectation, I dropped it. I would have given the 10k-digit problem more
I should also admit that I was a lot more lazy for p = (c*6*4691#+1). This is
a good sieve form and it turned out I needed 1314 primes (bad luck), but I
used pfgw's trial factoring for this. Later I discovered the NewPGen check box
"use primorial mode" in plain sight. I had only looked through the type
selection box... When it opens it actually covers the primorial box and I
thought there was only k*b^n something available. Why didn't Paul Jobling put
a hint in the type box for people like me with tunnel vision :-)
Never underestimate the potential for stupidity among software users.
Phil Carmody wrote:
> In the world of sieving, Jens is one of the people to whom I doff my hat.Thanks. However alt.math.recreational is not hard competition, at least before
> He stood out on alt.math.recreational because of his sieving ability.
you joined it. You are the better siever and I have only beaten your speed a
couple of times (including the prime puzzles) by spending much more time on
special-purpose programs. GenSv seems impressive for such a generic sieve.
Jens Kruse Andersen