20170306, 20:22  #1 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT)
1640_{16} Posts 
Runs of factored numbers
Been thinking this evening about factoring consecutive numbers.
Two numbers is easy: 2^74207281 and 2^742072811 Three numbers is also easy based upon the largest twin prime pair: 2996863034895 · 2^1290000 ± 1 and 2996863034895 · 2^1290000 Four or more numbers is a bit harder. Anyone got any ideas? It may be necessary to rely on prp cofactors beyond ECPP size. 
20170306, 20:59  #2 
"Robert Gerbicz"
Oct 2005
Hungary
2^{2}×3×5×23 Posts 
It is a nice problem, see here: http://www.math.uni.wroc.pl/~jwr/consfac/

20170306, 22:49  #3  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20170307, 03:59  #4 
Romulan Interpreter
Jun 2011
Thailand
21BB_{16} Posts 
You can take newpgen (or any other siever) and sieve 2^n+k for a VERY large particular n (pick one) and some k range like [1000, +1000], then see where are the gaps. Larger gap will give your series of consecutive numbers with factors, and the factors too.
Or were you talking about full factorization? Last fiddled with by LaurV on 20170307 at 04:02 
20170307, 04:44  #5 
Jun 2003
1001000111111_{2} Posts 

20170307, 17:57  #6 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT)
1640_{16} Posts 
It seems the idea used for the larger runs is for many of the numbers to have algebraic factors. The run of 20 numbers has 9/20 with algebraic factors splitting them in half.
I am not certain how to go about creating these polynomials or if it is possible to have a higher number with algebraic factors. 
20170307, 18:47  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×7×11×59 Posts 
length 22.. "new" record
Ok, I have just made it length 21 by factoring N1 :)
Code:
N1 = 5213411917...84<303> = 2^2 · 23 · 41 · 290399 · 82049473563797795110840675341962134165303 · 5800687119043966076458093431403599342066486860130593363750638259379043892237928868657375528720659036199535857003362322259881288749204326339939518541384047634456777775176259082126086907366473591808484877382252182189655036384810428841744360051529320492751 ...have just made it length 22 by factoring N2 :) Code:
N2 = 5213411917...83_{<303>} = 3 · 1231 · 2046425585967871763903675443<28> · 125250681966946964598302585902770629381_{<39>} · 5507654344...57_{<234>} Last fiddled with by Batalov on 20170307 at 23:25 
20170308, 14:46  #8  
Feb 2017
Nowhere
6420_{8} Posts 
Quote:
[Google, Google, toil and trouble] The expressions at Largest Consecutive Factorizations just below the first occurrence of "4178" (do a string search) yield a cornucopia of polynomial expressions involving curious substitutions. Hmm, looking at the references, A Chinese Prouhet–Tarry–Escott solution sort of stands out... BTW, Batalov has been credited with extending the run at the previouslyreferenced site. 

20170309, 19:24  #9  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT)
5696_{10} Posts 
Quote:


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