Timeline for Finding integer representation as difference of two triangular numbers
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 22, 2016 at 0:51 | comment | added | user25406 | This is not an answer but I am not allowed to comment yet ( and I apologize). my question is: can you please show how your method can factor the triangular number N=8*17=136? I tried finding two triangular numbers whose difference was N=136 and couldn't. I then tried to go backward from factors 8*17,2*68,4*34 to (a,b) setting a+b=2*17 and a-b+1=8...and I run into a problem. the only way to go backward is to use a+b=136 and a-b+1=1. And this may be interesting because it looks like triangular numbers have their own "primes". (I will delete my answer once your respond to my question) | |
| Dec 26, 2015 at 10:20 | vote | accept | CommunityBot | ||
| Dec 26, 2015 at 4:28 | answer | added | Jeremy Rouse | timeline score: 1 | |
| Dec 26, 2015 at 4:15 | answer | added | David Eppstein | timeline score: 3 | |
| Nov 6, 2015 at 20:11 | comment | added | user6671 | Every method to factor an uneven number is equivalent to Fermat's method: $2n+1=d\cdot e = ((d+e)/2)^2-((d-e)/2)^2$ | |
| Oct 26, 2015 at 15:45 | history | edited | user6671 | CC BY-SA 3.0 |
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| Oct 26, 2015 at 15:42 | comment | added | user6671 | I choose 1000 random pair of primes $p, q$ not equal each with 6 digits and measured the steps in the loop, for example: $p=108571,q=102329,n=11109961859,steps=7552$ In this case the exponent is $log(7552)/log(n) = 0.28631$ | |
| Oct 26, 2015 at 14:33 | comment | added | Bill Bradley | What range of $n$ did you examine to estimate your 0.38 exponent? | |
| Oct 26, 2015 at 14:25 | comment | added | user6671 | I know that, but I thought, that somebody might know of a faster way to find such representation. Thanks for your comment. | |
| Oct 26, 2015 at 14:22 | comment | added | Geoff Robinson | But what you are doing is equivalent to writing $8n = (2a+1)^{2}- (2b-1)^{2}$. | |
| Oct 26, 2015 at 14:13 | comment | added | user6671 | I know of Fermat's factorization method. This method is different. | |
| Oct 26, 2015 at 14:05 | comment | added | S. Carnahan♦ | This is just a minor variant of Fermat's method, but using triangular numbers instead of squares: en.wikipedia.org/wiki/Fermat%27s_factorization_method | |
| Oct 26, 2015 at 14:03 | answer | added | joro | timeline score: 0 | |
| Oct 26, 2015 at 13:35 | history | edited | user21574 |
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| Oct 26, 2015 at 11:39 | review | Close votes | |||
| Oct 26, 2015 at 18:35 | |||||
| Oct 26, 2015 at 11:10 | history | asked | user6671 | CC BY-SA 3.0 |