## 13 consecutive factorizations at 500 digits

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• I d like to report 13 consecutive factorizations at 500 digits. ... N = (2*x^3 - 3*x^2 - 5*x + 2)^2 * (4*x^6 - 12*x^5 - 11*x^4 + 38*x^3 + 13*x^2 - 20*x - 6)^2
Message 1 of 16 , Dec 15, 2009
I'd like to report 13 consecutive factorizations at 500 digits.
------------------------------
N = (2*x^3 - 3*x^2 - 5*x + 2)^2 * (4*x^6 - 12*x^5 - 11*x^4 + 38*x^3 + 13*x^2 - 20*x - 6)^2 / 16;
where x = 2^92+227683166;

N-k is factored for k={0..12}
------------------------------

All primes were proved using PARI/GP.

PARI/GP verification script available here:
http://immortaltheory.com/cnt/verify_k13.gp

As always my brother John Michael Crump helped with his ECM support! :)

Factorizations are abbreviated as follows:
---------------------------------------------
N-0 = 2^6 * 3^2 * 41^2 * 59^4 * 199^2 * 3907^2 * 1067282959^2 * p59^2 * p67^2 * p104^2
---------------------------------------------
N-1 = 5 * 7^2 * 19 * 2039 * 2203 * 18917 * 20773 * 128663 * 206197 * 337283 * 4205437 * 4253763823 * 10361106142511 * 48219445919633 * 38944767299226517 * 823146395611037243 * 2567947475640367903 * p31 * p46 * p105 * p189
---------------------------------------------
N-2 = 2 * 31 * 71 * 479 * 601 * 2521 * 10079 * 11351 * p34 * p35 * p43 * p46 * p94 * p112 * p119
---------------------------------------------
N-3 = 3 * 61 * 1019 * 14243 * 52883 * 269749 * 73255439 * 1462827239 * p22 * p37 * p44 * p90 * p125 * p147
---------------------------------------------
N-4 = 2^2 * 829 * 26342725653094583 * p28 * p33 * p53 * p63 * p88 * p217
---------------------------------------------
N-5 = 4001 * 33739 * p491
---------------------------------------------
N-6 = 2 * 3 * 5^2 * 29 * 2664932713 * 64053375527497 * p35 * p122 * p317
---------------------------------------------
N-7 = p500
---------------------------------------------
N-8 = 2^3 * 7 * 103 * 167 * 1823 * 9436961405541853927 * p67 * p75 * p330
---------------------------------------------
N-9 = 3^5 * 11 * 37 * 73 * 89 * 1697 * 22111 * 24379 * 50993 * 75403 * 294431 * 791899 * 7143163 * 37162403 * 40238771 * 318066373 * 774022187 * 1420502627 * 319986975727 * 360802459587989 * 4907888847800929 * p28 * p40 * p43 * p93 * p166
---------------------------------------------
N-10 = 2 * 13 * 227 * 1889 * p493
---------------------------------------------
N-11 = 5 * 5261 * p23 * p473
---------------------------------------------
N-12 = 2^2 * 3 * 1022689 * 11589211679 * p482
------------------------------

The ends, N+1 and N-13, were subjected to ~1700 ECM curves at B1=1e6 and ~500 curves at B1=3e6 leaving a c498 and c472.

PS: With another candidate x, I came within a c233 of finding a 14-streak. Unfortunately though that
one didn't factor after a considerable ECM effort. But suggests longer streaks are within range
with time and/or luck.
• CONGRATULATIONS !!! Impressive! What more can I say? My site http://www.math.uni.wroc.pl/~jwr/cons-fac/ is updated (since the results still fits Jens s limit
Message 2 of 16 , Dec 15, 2009
CONGRATULATIONS !!!

Impressive!

What more can I say?

My site
http://www.math.uni.wroc.pl/~jwr/cons-fac/
is updated (since the results still fits Jens's limit of 500 digits,
it will be listed primarily at his site).

Jarek

2009/12/16 Joe <joecr@...>
>
>
>
> I'd like to report 13 consecutive factorizations at 500 digits.
> ------------------------------
> N = (2*x^3 - 3*x^2 - 5*x + 2)^2 * (4*x^6 - 12*x^5 - 11*x^4 + 38*x^3 + 13*x^2 - 20*x - 6)^2 / 16;
> where x = 2^92+227683166;
>
> N-k is factored for k={0..12}
> ------------------------------
>
> All primes were proved using PARI/GP.
>
> PARI/GP verification script available here:
> http://immortaltheory.com/cnt/verify_k13.gp
>
> As always my brother John Michael Crump helped with his ECM support! :)
>
> Factorizations are abbreviated as follows:
> ---------------------------------------------
> N-0 = 2^6 * 3^2 * 41^2 * 59^4 * 199^2 * 3907^2 * 1067282959^2 * p59^2 * p67^2 * p104^2
> ---------------------------------------------
> N-1 = 5 * 7^2 * 19 * 2039 * 2203 * 18917 * 20773 * 128663 * 206197 * 337283 * 4205437 * 4253763823 * 10361106142511 * 48219445919633 * 38944767299226517 * 823146395611037243 * 2567947475640367903 * p31 * p46 * p105 * p189
> ---------------------------------------------
> N-2 = 2 * 31 * 71 * 479 * 601 * 2521 * 10079 * 11351 * p34 * p35 * p43 * p46 * p94 * p112 * p119
> ---------------------------------------------
> N-3 = 3 * 61 * 1019 * 14243 * 52883 * 269749 * 73255439 * 1462827239 * p22 * p37 * p44 * p90 * p125 * p147
> ---------------------------------------------
> N-4 = 2^2 * 829 * 26342725653094583 * p28 * p33 * p53 * p63 * p88 * p217
> ---------------------------------------------
> N-5 = 4001 * 33739 * p491
> ---------------------------------------------
> N-6 = 2 * 3 * 5^2 * 29 * 2664932713 * 64053375527497 * p35 * p122 * p317
> ---------------------------------------------
> N-7 = p500
> ---------------------------------------------
> N-8 = 2^3 * 7 * 103 * 167 * 1823 * 9436961405541853927 * p67 * p75 * p330
> ---------------------------------------------
> N-9 = 3^5 * 11 * 37 * 73 * 89 * 1697 * 22111 * 24379 * 50993 * 75403 * 294431 * 791899 * 7143163 * 37162403 * 40238771 * 318066373 * 774022187 * 1420502627 * 319986975727 * 360802459587989 * 4907888847800929 * p28 * p40 * p43 * p93 * p166
> ---------------------------------------------
> N-10 = 2 * 13 * 227 * 1889 * p493
> ---------------------------------------------
> N-11 = 5 * 5261 * p23 * p473
> ---------------------------------------------
> N-12 = 2^2 * 3 * 1022689 * 11589211679 * p482
> ------------------------------
>
> The ends, N+1 and N-13, were subjected to ~1700 ECM curves at B1=1e6 and ~500 curves at B1=3e6 leaving a c498 and c472.
>
> PS: With another candidate x, I came within a c233 of finding a 14-streak. Unfortunately though that
> one didn't factor after a considerable ECM effort. But suggests longer streaks are within range
> with time and/or luck.
>
>
• ... Congratulations on the first 13! I m impressed. The length record has now increased by 3 in 18 days.
Message 3 of 16 , Dec 16, 2009
Joe wrote:
> I'd like to report 13 consecutive factorizations at 500 digits.
> ------------------------------
> N = (2*x^3 - 3*x^2 - 5*x + 2)^2 * (4*x^6 - 12*x^5 - 11*x^4
> + 38*x^3 + 13*x^2 - 20*x - 6)^2 / 16;
> where x = 2^92+227683166;
>
> N-k is factored for k={0..12}

Congratulations on the first 13!
I'm impressed.
The length record has now increased by 3 in 18 days.
http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm
is updated.

> The ends, N+1 and N-13, were subjected to ~1700 ECM curves
> at B1=1e6 and ~500 curves at B1=3e6 leaving a c498 and c472.

You didn't mention the prime factors but it was easy to find:
N+1 = 17 * c498
N-13 = 7909460491591 * 854342030873171 * c472

--
Jens Kruse Andersen
• ... thus neatly satisfying both Jens and Jarek. ... 17, perhaps? {N=(2*x^3-3*x^2-5*x+2)^2 *(4*x^6-12*x^5-11*x^4+38*x^3+13*x^2-20*x-6)^2/16;}
Message 4 of 16 , Dec 16, 2009
--- In primeform@yahoogroups.com,
"Joe" <joecr@...> wrote:

> 13 consecutive factorizations at 500 digits

thus neatly satisfying both Jens and Jarek.

> longer streaks are within range

17, perhaps?

{N=(2*x^3-3*x^2-5*x+2)^2
*(4*x^6-12*x^5-11*x^4+38*x^3+13*x^2-20*x-6)^2/16;}
{for(k=0,16,f=factor(N-k)[,1];if(#f>1,
print([k,vector(#f,j,poldegree(f[j]))])))}
[0, [3, 6]]
[1, [3, 6, 9]]
[2, [2, 4, 6, 6]]
[3, [6, 6, 6]]
[4, [3, 6, 9]]
[6, [6, 12]]
[8, [6, 12]]
[9, [1, 1, 1, 3, 6, 6]]
[14, [6, 12]]
[16, [9, 9]]

All the best!

David
• Factorization of 16 consecutive 303-digit integers Let x = 9*(2^123 + 23815893) N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 then for k = 0
Message 5 of 16 , Dec 23, 2009
Factorization of 16 consecutive 303-digit integers

Let x = 9*(2^123 + 23815893)
N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216
then for k = 0 to 15 the factorization of N+k is in
I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
It is unlikely that either of the composite integers
(N-1)/(3*5*107614765903*707599683401162722538492715491)
(N+16)/(2*59*347*509*28099*71147*63442396558253*535991020920557)
is divisible by a prime less than 10^35.

• Congratulations!!! My site http://www.math.uni.wroc.pl/~jwr/cons-fac/ is updated. Jarek 2009/12/24 djbroadhurst ... [Non-text
Message 6 of 16 , Dec 23, 2009
Congratulations!!!

My site
http://www.math.uni.wroc.pl/~jwr/cons-fac/
is updated.

Jarek

>
>
>
>
> Factorization of 16 consecutive 303-digit integers
>
> Let x = 9*(2^123 + 23815893)
> N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216
> then for k = 0 to 15 the factorization of N+k is in
> I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
> It is unlikely that either of the composite integers
> (N-1)/(3*5*107614765903*707599683401162722538492715491)
> (N+16)/(2*59*347*509*28099*71147*63442396558253*535991020920557)
> is divisible by a prime less than 10^35.
>
> David Broadhurst, 24 December, 2009
>
>
>

[Non-text portions of this message have been removed]
• ... You might add that N-1 is divisible by p40 = 5791987309280956264759051605269253758123 David
Message 7 of 16 , Dec 24, 2009
--- In primeform@yahoogroups.com,
Jaroslaw Wroblewski <Jaroslaw.Wroblewski@...> wrote:

> My site
> http://www.math.uni.wroc.pl/~jwr/cons-fac/
> is updated.

You might add that N-1 is divisible by
p40 = 5791987309280956264759051605269253758123

David
• 2009/12/24 djbroadhurst ... Done. That leaves c221 to factor. Jarek [Non-text portions of this message have been removed]
Message 8 of 16 , Dec 24, 2009

> You might add that N-1 is divisible by
> p40 = 5791987309280956264759051605269253758123
>

Done. That leaves c221 to factor.

Jarek

[Non-text portions of this message have been removed]
• Happy X-Mas to you , primefans ! best wishes Norman __________________________________________________ Do You Yahoo!? Sie sind Spam leid? Yahoo! Mail verfügt
Message 9 of 16 , Dec 24, 2009
Happy X-Mas to you , primefans !

best wishes

Norman

__________________________________________________
Do You Yahoo!?
Sie sind Spam leid? Yahoo! Mail verfügt über einen herausragenden Schutz gegen Massenmails.
http://mail.yahoo.com

[Non-text portions of this message have been removed]
• Thanks, you too !!! - Joe Crump ... From: Norman Luhn To: primeform@yahoogroups.com Sent: Thursday, December 24, 2009 1:55 PM Subject: [primeform] 2,3,5,7,....
Message 10 of 16 , Dec 24, 2009
Thanks, you too !!!

- Joe Crump

----- Original Message -----
From: Norman Luhn
To: primeform@yahoogroups.com
Sent: Thursday, December 24, 2009 1:55 PM
Subject: [primeform] 2,3,5,7,....

Happy X-Mas to you , primefans !

best wishes

Norman

__________________________________________________
Do You Yahoo!?
Sie sind Spam leid? Yahoo! Mail verfügt über einen herausragenden Schutz gegen Massenmails.
http://mail.yahoo.com

[Non-text portions of this message have been removed]

[Non-text portions of this message have been removed]
• Factorization of 18 consecutive 303-digit integers Let x = 9*(2^123 + 63841119); N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216; then for k =
Message 11 of 16 , Dec 26, 2009
Factorization of 18 consecutive 303-digit integers

Let x = 9*(2^123 + 63841119);
N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216;
then for k = 0 to 17 the factorization of N+k is in
I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
It is unlikely that either of the composite integers
(N-1)/(3^2*7660729*75495982961295157)
(N+18)/(2*811*2777*562633*398311899246763)
is divisible by a prime less than 10^35.

• Congratulations!!! http://www.math.uni.wroc.pl/~jwr/cons-fac/ is updated. Can t wait to see if a 20 is possible to find :-))) Jarek 2009/12/27 djbroadhurst
Message 12 of 16 , Dec 26, 2009
Congratulations!!!

http://www.math.uni.wroc.pl/~jwr/cons-fac/
is updated.

Can't wait to see if a 20 is possible to find :-)))

Jarek

>
>
> Factorization of 18 consecutive 303-digit integers
>
> Let x = 9*(2^123 + 63841119);
> N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216;
> then for k = 0 to 17 the factorization of N+k is in
> I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
> It is unlikely that either of the composite integers
> (N-1)/(3^2*7660729*75495982961295157)
> (N+18)/(2*811*2777*562633*398311899246763)
> is divisible by a prime less than 10^35.
>
> David Broadhurst, 27 December, 2009
>
>
>

[Non-text portions of this message have been removed]
• Factorization of 20 consecutive 303-digit integers Let x = 9*(2^123 + 440959466) N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 - 1 then for k
Message 13 of 16 , Jan 4, 2010
Factorization of 20 consecutive 303-digit integers

Let x = 9*(2^123 + 440959466)
N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 - 1
then for k = 0 to 19 the factorization of N+k is in
I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
It is unlikely that either of the composite integers
(N-1)/(2^2*23*41*290399)
(N+20)/(5*19*205889393657*301133059382193691*57657649694534487023)
is divisible by a prime less than 10^40.

• ... Congratulations on reaching this mark! -- Jens Kruse Andersen
Message 14 of 16 , Jan 4, 2010
> Let x = 9*(2^123 + 440959466)
> N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 - 1
> then for k = 0 to 19 the factorization of N+k is in

30 November 2009 Jaroslaw Wroblewski wrote:
> I estimate that at 300 digits the longest submissions
> could be around 20 numbers.

Congratulations on reaching this mark!

--
Jens Kruse Andersen
• ... It was quite tough: only 9/20 came for free; getting the other 11 from OpenPFGW and GMP-ECM was not easy. In fact, I did not find a run of 19 before this
Message 15 of 16 , Jan 4, 2010
--- In primeform@yahoogroups.com,
"Jens Kruse Andersen" <jens.k.a@...> wrote:

> Congratulations on reaching this mark!

other 11 from OpenPFGW and GMP-ECM was not easy. In fact,
I did not find a run of 19 before this run of 20, which
came from plugging a hole by extracting
p47 = 21736936592318538144041717556296066161071377743
from
(N+2)/(43*7537*23671) = c293 = p47*p247

David
• BIG CONGRATULATIONS !!! I am impressed !!! My site http://www.math.uni.wroc.pl/~jwr/cons-fac/ is updated. Jarek 2010/1/5 djbroadhurst
Message 16 of 16 , Jan 4, 2010
BIG CONGRATULATIONS !!!

I am impressed !!!

My site
http://www.math.uni.wroc.pl/~jwr/cons-fac/
is updated.

Jarek

>
>
>
>
> Factorization of 20 consecutive 303-digit integers
>
> Let x = 9*(2^123 + 440959466)
> N = (x-2)*(2*x+9)*(2*x^2+3*x-8)*(2*x^2+5*x+5)*(2*x^2+7*x-3)/216 - 1
> then for k = 0 to 19 the factorization of N+k is in
> I used OpenPFGW, GMP-ECM, Msieve, GGNFS and Pari-GP.
> It is unlikely that either of the composite integers
> (N-1)/(2^2*23*41*290399)
> (N+20)/(5*19*205889393657*301133059382193691*57657649694534487023)
> is divisible by a prime less than 10^40.
>
> David Broadhurst, 5 January, 2010
>
>
>

[Non-text portions of this message have been removed]
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