mersenneforum.org Lepore Factorization in O(k) Conjecture
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 2017-08-30, 10:57 #1 Alberico Lepore     May 2017 ITALY 2·32·29 Posts Lepore Factorization in O(k) Conjecture Lepore Factorization in O(k) Conjecture https://www.academia.edu/34408457/Le...O_k_Conjecture Do you think about it?
 2017-08-30, 16:28 #2 CRGreathouse     Aug 2006 22×3×499 Posts I don't see either a conjecture or an algorithm there. But I have a conjecture of my own: any paper which spends more than half its length exploring the consequences of numbers being of the form 6n ± 1 is unlikely to make an algorithmic or number-theoretic breakthrough. I know my conjecture seems oddly specific, but I've seen it pretty often. (At least with half replaced with a more reasonable fraction like a tenth.)
2017-08-31, 00:19   #3
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

23×439 Posts

Quote:
Quote:
 Page Not Found Sorry. We can't find the page you're looking for.
In addition to Charles' conjecture, here's another one:

Any paper that has the author's name also in the title is not worth reading.
(e.g. J.Kruasandwich, "The Kruasandwich method of turning lead into gold". Stop right there. Don't read it.)

 2017-09-03, 09:26 #4 Alberico Lepore     May 2017 ITALY 2×32×29 Posts Look this https://www.academia.edu/34339494/Fa...t%C3%A0_random And think this If we go to look directly, numbers with k
2017-09-03, 10:24   #5
jnml

Feb 2012
Prague, Czech Republ

3118 Posts

Quote:
 Originally Posted by Alberico Lepore Look this https://www.academia.edu/34339494/Fa...t%C3%A0_random And think this If we go to look directly, numbers with k
PSA: Let me advise anyone against downloading the paper for at least two reasons: it has authors name in its title (cf. Batalov's post above) and downloading it requires to "connect" with the author on G+ or FB. Would the intents of the author be fair, there're zero reasons to hide the paper behind any paywall/"connectwall" etc.

And if you do download the paper and you're using a popular unsecure OS, do not open the document without performing proper AV checks.

Last fiddled with by jnml on 2017-09-03 at 10:25 Reason: s/reazons/reasons/, s/it's/its/

 2017-09-03, 10:27 #6 Alberico Lepore     May 2017 ITALY 20A16 Posts Fattorizzazione RSA di Lepore Complessità random Definizione Tutti i numeri NR escluso i multipli di 2 e di 3 si scrivono nella forma 6h+1 e 6h+5. Dimostrazione NR modulo 6 =1 -> 6h+1 NR modulo 6 =2 -> è multiplo di 2 NR modulo 6 =3 -> è multiplo di 3 NR modulo 6 =4 -> è multiplo di 2 NR modulo 6 =5 -> 6h+5 NR modulo 6 =0 -> è multiplo di 2 e di 3 Lemma Quindi partendo da 1 e facendo +4 e +2 si ha 1 5 7 11 13 17 19 23 25 29 ecc.ecc. Definizione Ogni numero NR escluso i multipli di 2 e di 3 si scrivono nella forma 1) X^2+6nX=NR 2) X^2+6nX+2X=NR 3) X^2+6nX+4X=NR Dimostrazione Dal lemma segue direttamente 1) X(X+6n)=NR 2) X(X+6n+2)=NR 3) X(X+6n+4)=NR In più si può osservare che: (6h+1)*(6k+1)=6G+1 (6h+5)*(6k+5)=6G+1 (6h+1)*(6k+5)=6G+5 (6h+5)*(6k+1)=6G+5 ---------------------------------------------------------------------------------------- Da ciò si può dedurre che risolvendo (6h+1)*(6k+1)=6G+1 si possano risolvere gli altri tre casi questi (6h+1)*(6k+5)=6G+5 (6h+5)*(6k+1)=6G+5 moltiplicando per 5 e questo (6h+5)*(6k+5)=6G+1 moltiplicando per 25 Quindi prendiamo come caso base (6h+1)*(6k+1)=6G+1 X^2+6nX=NR si può facilmente notare che se G è pari n sarà pari se G è dispari n sarà dispari ---------------------------------------------------------------------------------------- Se NR=(6*a+1)*(6*b+1) allora possiamo scriverlo nella forma NR=(6*a+1)^2+6*n*(6*a+1) quindi n=(G-6*a^2-2*a)/(6*a+1) dove G=(NR-1)/6 ---------------------------------------------------------------------------------------- Per capire ci scriviamo una tabella dove i valori NR=(6*a+1)^2+6*n*(6*a+1) a parte da 1 ed n parte da 0 a\n 49 91 133 175 217 259 …....... 169 247 325 403 481 559 …...... 361 475 589 703 817 931 …...... 625 775 925 1075 1225 1375 …...... 961 1147 1333 1519 1705 1891 …....... 1369 1591 1813 2035 2257 …................ …............................................................... …............................................................... Si può osservare che la differenza tra NR(a+1,n-2)-NR(a,n)=36*(n-1) Quindi l'idea è di aggiungere ad NR un multiplo di 36 Cioè NR2=NR+i*36 fino a quando non soddisfi una determinata condizione cioè questa NR=(6*a+1)^2+6*n*(6*a+1) con (6*a+1) divisore di NR calcolandoci la n dalle seguenti da una delle due Se (NR-1)/6 è dispari (2+2*N)*N/2-(2+2*M)*M/2=(NR2-NR1)/36=K dove n=2*N+1 Se (NR-1)/6 è pari N^2-M^2=(NR2-NR1)/36=K dove n=2*N [(*NOTA1*) per il momento tralascio come si calcola n ed m , ci devo pensare un po] Quindi avremo n-1 possibilità di risolvere la fattorizzazione. Da notare che più grande è i più la frequenza dei numeri che ci interessano è bassa. Esempi [(*NOTA2*) In questi esempi terremo conto che di solito in RSA il numero da fattorizzare NR=p*q avrà q/p < 2 ] Esempio 1 617251=p*q=p^2+6*n*p per la (*NOTA2*) p >= 553 e q <= 1111 quindi la n massima è n_max=(q-p)/6=93 i numeri i validi sono 84 (2+2*N)*N/2-(2+2*M)*M/2=84 N=42 n=2*42+1=85 617251=p^2+6*85*p segue p=571 166 (2+2*N)*N/2-(2+2*M)*M/2=166 N=42 n=2*42+1=85 246 (2+2*N)*N/2-(2+2*M)*M/2=246 N=42 n=2*42+1=85 324 400 474 546 ecc. ecc. Esempio 2 620677=p*q=p^2+6*n*p i numeri i validi sono 85 N^2-M^2=85 N=43 n=2*43=86 620677=p^2+6*86*p segue p=571 ecc. ecc. Alberico Lepore 24 Agosto 2017
 2017-09-05, 03:54 #7 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 282D16 Posts Wow! A wonderful algorithm. Can you please factor the following C154 number for me? Code: 9244198061795738171803076086524911379752268769558126772679104427875162378559757802686251958103938360845772387320740454715805477545200393208224631568651017 If you find the factors fast, you save me the time to find a poly for it, to do sieving, filtering, LA, etc. With O(k) you will only need few hours, instead of me spending a week or so... Thanks in advance. P.S. If you can not, then get the heck out of here, and stop polluting the forum with rubbish... Last fiddled with by LaurV on 2017-09-05 at 03:57
 2017-09-05, 19:43 #8 Max0526   "Max" Jun 2016 Toronto 929 Posts C154 poly? @LaurV On a serious note, do you need a (good) poly for this C154?
2017-09-06, 13:13   #9
Dr Sardonicus

Feb 2017
Nowhere

11000110000002 Posts

Quote:
 Originally Posted by jnml PSA: Let me advise anyone against downloading the paper for at least two reasons: it has authors name in its title (cf. Batalov's post above) and downloading it requires to "connect" with the author on G+ or FB.
My reasoning was even briefer: The conjecture wasn't stated at the link where the post said it was. Therefore, as far as I was concerned, the poster was up to no good. I wasn't about to download a file just to find out what his purported conjecture was. Besides -- what was stated in the link given in the post seemed not to be leading anywhere interesting.

And so it is that I remain happily in ignorance of what the "k" signifies in the "O(k)" part of the purported conjecture

2017-09-06, 14:29   #10
CRGreathouse

Aug 2006

22×3×499 Posts

Quote:
 Originally Posted by Dr Sardonicus And so it is that I remain happily in ignorance of what the "k" signifies in the "O(k)" part of the purported conjecture
Likewise, despite having read the material.

 2017-09-06, 14:34 #11 firejuggler     "Vincent" Apr 2010 Over the rainbow 22×7×103 Posts you know, k,n,m,p x,y,z...... different country have different notation for the same think... log(n) , log(y), or sin(n) shouldn't phase you.

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