20190407, 16:35  #34 
Nov 2005
101 Posts 
Sorry.
The sum of all divisors  not only the prime divisors I thought of  can be higher then the number itself. So I misinterpreted your model. So yes abundant numbers https://en.wikipedia.org/wiki/Abundant_number are good candidates. But for example 105 and 15 are no such numbers but are good 'k'. Last fiddled with by ThiloHarich on 20190407 at 17:01 
20190407, 17:03  #35  
Apr 2019
40_{8} Posts 
Quote:
Last fiddled with by nesio on 20190407 at 17:11 Reason: grammar correction 

20190407, 18:16  #36  
"Tilman Neumann"
Jan 2016
Germany
2^{2}·109 Posts 
We have some language problems here. Just to clarify: 12 has divisors 1, 2, 3, 4, 6, 12. Is 6 the result you mean, or is it 28? The rough model is the one he proposed, not yours: Quote:
The quality of a multiplier m is about S/(m^1/3), where we consider m < n^1/3 only. S might be a) the sum of divisors of m, or b) the number of disivors of m. My first question in this post aims to clarify which of the two options is the correct one. nesio, do I get that right so far? Last fiddled with by Till on 20190407 at 18:19 Reason: make clear that only one option for S is correct 

20190407, 18:31  #37 
Apr 2019
2^{5} Posts 
A rough model of m:
Find the maximum of (S/(m^1/3)) when m < n^1/3, where S  the number of all divisors of m (excluding 1 and m). Here are S and m^1/3 have equal weights in maximize function. 
20190407, 18:43  #38  
"Tilman Neumann"
Jan 2016
Germany
664_{8} Posts 
Ok, now we understand that one. Quote:
I'ld say this is not a model of m, but a procedure to choose the best m. A model would give a kind of score to any m. (at least) A simple weighting function giving a "plus" for smooth m, a "minus" for big m. This is no rocket science. We tested dozens of ways to arrange k's according to such principles, and many other constellations... 

20190407, 19:04  #39 
Apr 2019
32_{10} Posts 
Is it working on your data and what are the results?
Last fiddled with by nesio on 20190407 at 19:11 Reason: lang 
20190407, 19:39  #40 
"Tilman Neumann"
Jan 2016
Germany
2^{2}·109 Posts 
Unfortunately, none of our attempts to sort k's by such measures worked well.
I believe that you did not try an efficient implementation yet. Theory (aka computational complexity, counting iterations being a simplified version of) and reality (measuring speed) are quite different for "small" factoring algorithms. After all, this is the only reason why Lehman or Hart algorithms have a right to exist. Otherwise NFS would wipe out all other algorithms at any number scale. If you try to optimize Hart/Lehman for small n, say n<=60 bit, then you will realize that instructions that are very fast on that number scale have a big or even decisive impact on the general performance. Complexity theory is often not strong enough to capture such influences. Storing the sqrts of k or of km is such a point. Without it, you will not be able to get a really fast Hart/Lehman implementation. Using congruences is a theoretical improvement, but also necessary. You can disprove my assumption that you did not try an efficient implementation yet by posting a fast piece of software ;) 
20190407, 21:29  #41 
Apr 2019
2^{5} Posts 
For example such rough model (procedure) gives these values m from n
n m 555 6 5555 12 55555 36 555555 60 5555555 120 55555555 360 555555555 720 5555555555 1680 55555555555 2520 555555555555 5040 5555555555555 5040 55555555555555 5040 555555555555555 5040 It is light and quite workable way for m in "multiplication method" we hope. Til! Our work is a research and a paper about Recursive Multiplication algorithm and its relation to Simple Multiplication algorithm. Your work is improving Lehman and Harts algorithms. You asked us about our vision regarding m. We wrote then. Something does not suit you in our discussion? Let's not continue to cross the red and wet ;) We think you and Thilo should write a paper about your work that may be interesting not only in that forum. Last fiddled with by nesio on 20190407 at 21:30 Reason: lang 
20190408, 06:18  #42 
"Tilman Neumann"
Jan 2016
Germany
2^{2}·109 Posts 
Hi nesio,
sorry if I have been unpatient. I do not understand Russian, so I cannot get all information from your paper. And it took some time to get some interesting information from you. I will try to implement your particular proposition of sorting m by Code:
score(m) = S(m) / n^(1/3) with S(m) = sum of divisors(m) without 1 and m Maybe we made an error so far in our evaluations of the smoothness of m's. We will see. You said that the computation of score(m) above is a very rough model. What would a more elaborate model look like? 
20190408, 06:39  #43 
If I May
"Chris Halsall"
Sep 2002
Barbados
9577_{10} Posts 

20190408, 09:27  #44 
Nov 2005
101 Posts 
 Choosing the right m depends on the implementation.
Til and me used some adjustments/improvements for odd multipliers m > in our algorithm the good multipliers are not only even, best multipliers are odd. the numbers from nesio model : 6,12,36,60,120,360,720,1680,... do not appear in the list of good multipliers for this tuned algorithm in the top  the number of divisors is the product of the exponents + 1 of its prime factors If the exponents are all 1 then this is just 2^(number of prime factors of m). Having a factor with an exponent greater then one reduces the number of divisors. > it looks like a good measure for good multipliers. But our best multiplier 315 = 3^2 * 5 * 7 has an exponent 2. When finding the best multipliers for numbers starting at 2^54 (and only factoring numbers with factors above n^1/3) I get a complete different list then when factoring numbers starting at 2^55. Last fiddled with by ThiloHarich on 20190408 at 09:33 
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