20161211, 05:34  #1 
"Sam"
Nov 2016
5×67 Posts 
Prime Gap Length with consecutive integers divisible by small primes
I see this subforum is about prime gaps, and a long list of consecutive composites can be obtained by restricting that each number from n to n+k has at least one prime factor < s where s is the factor bound.
For instance there are at most 5 consecutive integers such that each of them is divisible by 2, 3, or 5. There at most 9 consecutive integers such that each of them is divisible by 2, 3, 5, or 7. This pattern continues here. Does someone know the modular congruence for 257 consecutive integers, each of them is divisible by at least one prime less than 100? Thanks if so. Here, there is a 159 "minimum" prime gap, since each integer from n to n+158 is divisible by either 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, or 97. n = 1447667319088187212520359882964797112 According to the Oeis sequence I found, there are at least 98 more integers that can be divisible by at least one of the primes < 100 (257 total). I could have written out the modular congruence for the example I gave, but it would take too long, so I used factordb.com instead. 
20161211, 18:13  #2  
Jun 2003
Suva, Fiji
2^{3}·3·5·17 Posts 
Quote:
I don't know about 257  recently I looked at a gap of 2310 with 113 primes providing cover and the results of that were posted at http://www.mersenneforum.org/showthread.php?t=20825 

20161211, 18:55  #3  
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 
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20161211, 20:20  #4 
"Sam"
Nov 2016
5×67 Posts 
Sm88, I formed my example above my matching numbers that have not been "eliminated" with other factors I have not used yet. For example,
let (2*3*5*7*13*17*19*29*31*37*41*43*47*53*59*61*71*73*79*83*97)  n 11  n+1 Now each number k that is 1 (mod 11), 11  n+k. In this case, n+{23, 67, 89} are all divisible by 11, so making each of them (23, 67, and 89) not divide n "saves" prime numbers we could align later with numbers that don't have a specified factor < 100. If n1  23 and n+11  67, n11  89 or n+11  89, n11  67 There are now at least 123 consecutive integers such that each is divisible by a prime < 100. Here n22 = 1962136467370308094802867946933361058. Last fiddled with by carpetpool on 20161211 at 20:41 
20161211, 21:13  #5  
"Forget I exist"
Jul 2009
Dumbassville
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20161211, 21:42  #6 
"Sam"
Nov 2016
5×67 Posts 
Things are simpler when dealing with n# (product of all primes less than n). Just remove one prime factor q from n# (but not 2) and divide n# by all the prime congruent to 1 (mod q). You can use all the primes 1 (mod q) to align with other numbers which don't have a specified factor < 100, which is what I did.
Another alternative is by trial and error: choose some starting value n and find a prime factor < 100 for all numbers n, n+1, n+2, n+3,..... n+200. 2  n, and 2  n+k if k is even Find the number of integers relatively prime to 2 less than 200, and take each one mod 3. Take the modulo result that occurs the most (either 0, 1, or 2) and which ever one it is (actually it is 1) align 3 as a factor for that specific modular group. Eg. 3  n+1, and 3  n+k for all k = 1 (mod 3). Do this with 5 (find all the numbers relatively prime to 6 less than 200 and take each one mod 5, and find the mode m, of the modular results) Align n+m  5. And so on. I find this approach practically useless in some cases because the previous modular congruence will affect the next prime's congruence and so on. There are 2305567963945518424753102147331756070 to which all the prime factors can be aligned, about half of these combinations are impossible because of the congruence divisibility by 2. 
20161211, 22:22  #7  
"Forget I exist"
Jul 2009
Dumbassville
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20161212, 02:33  #8 
"Sam"
Nov 2016
5·67 Posts 
I use perl ntheory module for that: https://metacpan.org/pod/ntheory

20161212, 02:40  #9  
Aug 2006
5,987 Posts 
Quote:


20161212, 02:41  #10  
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 
Quote:
Last fiddled with by science_man_88 on 20161212 at 02:41 

20161212, 04:43  #11 
"Sam"
Nov 2016
5×67 Posts 
FYI I have Pari/GP too. However the perl ntheory module uses less memory on my computer than the Pari/GP interface so I use that more often.
Last fiddled with by carpetpool on 20161212 at 04:43 
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