

Thread Tools 
20170625, 18:44  #166  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}·269 Posts 
Quote:
It is not theoretically impossible but very impractical. Last fiddled with by a1call on 20170625 at 19:03 

20170625, 19:43  #167  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}·269 Posts 
Quote:
21/9 > 63/9 > 21/3. Of course unless the scale has a common factor grater than 1 with m, then it's not of much use. As a measure of potential complexity and extra numbers of steps involved in geometrical solutions versus algebraical solutions, consider the solution I came up with for 5 inscribed circles in a circle here: http://www.mersenneforum.org/showthread.php?t=21141 vs. its algebraic solution: Code:
The trigonometric formula for the radius r1 of n equal, tangent and inscribed circles inside a larger circle of radius r2 is: r1 = (r2 sin (180/n))/(1+sin (180/n)) Last fiddled with by a1call on 20170625 at 19:59 

20170625, 20:36  #168 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}·269 Posts 
This might be one theoretical use of the concept:
Code:
theExp = 5;\\larger integers will enlarge random n # \\\ primorial(n)= { my(thePrimorial=1); forprime(p=2,n,thePrimorial=thePrimorial*p); return (thePrimorial); } \\\ n=nextprime(random(19^theExp ))*nextprime(random(19^theExp )) factor(n) ## pr= primorial(sqrtint(n)); ## { s= n/ pr; print("** ",numerator(s)) } ## Code:
(16:31) gp > n=nextprime(random(19^theExp ))*nextprime(random(19^theExp )) %10 = 5452058249203 (16:31) gp > factor(n) %11 = [2236807 1] [2437429 1] (16:31) gp > ## *** last result computed in 0 ms. (16:31) gp > (16:31) gp > pr= primorial(sqrtint(n)); time = 7,376 ms. (16:31) gp > ## *** last result computed in 7,376 ms. (16:31) gp > { s= n/ pr; print("** ",numerator(s)) } ** 2437429 (16:31) gp > ## *** last result computed in 0 ms. (16:31) gp > 
20170625, 22:06  #169 
Aug 2006
13533_{8} Posts 
As a sidenote, might I suggest
Code:
n=randomprime([2,19^5])*randomprime([2,19^5]); Code:
theExp = 5;\\larger integers will enlarge random n n=nextprime(random(19^theExp ))*nextprime(random(19^theExp )) That was you get a uniform selection* and it's a little easier to see what's going on. Also, you should probably increase the lower bound  maybe randomprime([1e5, 3e5]) would be better. * Consider that, with theExp = 5, 2010881 is chosen 148 times more often than 3 and 74 times more often than 2010973. 
20170625, 22:29  #170 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}×269 Posts 
Cool.
I did not know that, Thank you. 
20170625, 23:07  #171 
Aug 2006
3×1,993 Posts 
I posted a very fiddly program
Code:
rsp(len)=my(mn=10^(len1),mx=10^len1,p=randomprime([sqrtup(mn/2),sqrtint(mx)]),q=randomprime([ceil(mn/p),mx\p])); p*q; Last fiddled with by CRGreathouse on 20170625 at 23:07 
20170625, 23:20  #172 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}×269 Posts 
My version doesn't seem to recognize sqrtup()
works fine with ceil(sqrt(mn/2)), though. Cool code, thank you. Adding it to my library under semiprimes. 
20170626, 00:04  #173  
Aug 2006
175B_{16} Posts 
Quote:
Code:
sqrtup(n)=n=ceil(n); if(issquare(n,&n),n,sqrtint(n)+1); Last fiddled with by CRGreathouse on 20170626 at 00:05 

20170626, 00:17  #174 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}·269 Posts 
Got it.
Thank you very much. 
20170627, 18:50  #175  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}·3·5 Posts 
Quote:
And in fact, if one used the expanded squares of a given representation, one can find more factors. Take the example of the first 4sq rep (5,5,5,4). 5^2=4^2+2^2+2^2+1^2 5^2=4^2+2^2+2^2+1^2 5^2=4^2+2^2+2^2+1^2 4^2=2^2+2^2+2^2+2^2 combining these squares, it is easy to see that: 4^2 + 2^2 + 1^2 = 21 = 3*7 2^2 + 2^2 + 2^2 + 1^1 = 13 2^2 + 1^2 + 1^2 + 1^2 = 7 4^2 + 2^2 + 2^2 + 2^2 = 28 = 4*7 4^2 + 4^2 + 1^2 + 1^2 + 1^2 = 35 = 5*7 .... 

20170627, 21:40  #176  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
9 8,1 7,2 7,1,1 6,3 6,2,1 6,1,1,1 5,4 5,3,1 5,2,2 5,2,1,1 5,1,1,1,1 4,5 4,4,1 4,3,2 4,3,1,1 4,2,1,1,1 4,1,1,1,1 3,6 3,5,1 3,4,2 3,4,1,1 3,3,1,1,1 3,2,1,1,1,1 3,1,1,1,1,1,1 3,3,3 3,3,2,1 3,3,1,1,1 3,2,1,1,1,1 3,1,1,1,1,1,1 ... all the red are partitions of another partition. some of these are also duplicates of earlier ones 3,4,2 and 4,3,2 and 2,3,4 for example ( there's actually 6 possible ways to rearrange these but that's also why you can cut the work back searching for the square sums because there are 16 with sign changes, each having 24 orderings. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A Sierpinski/Riesellike problem  sweety439  sweety439  1250  20211001 12:53 
Another factoring method rides the bus  3.14159  Factoring  233  20110515 18:50 
Square numbers and binary representation  ET_  Miscellaneous Math  40  20100606 12:55 
Is this a feasible factoring method?  1260  Miscellaneous Math  46  20100531 17:37 
Representation of integer as sum of squares  kpatsak  Math  4  20071029 17:53 