20171226, 21:31  #34 
Aug 2006
3·1,993 Posts 

20171226, 22:00  #35  
May 2017
ITALY
2×3^{2}×29 Posts 
Quote:
it does not depend on the number of digits but by the composition of the number. [1490166172*a^2]=a*(an)=Z , Z2*(an)^2=an , a^2+n*a=149016617 Time 2^1 67586227 Time 2^11 1500000021500000077 Time 2^2 [15000000215000000772*a^2]=Z , Z=a*(an), Z2*(an)^2=(an)*[(an)(n)]=Q , Q2*[(an)(n)]^2=(an)(n)(n) , a^2+n*a=1500000021500000077 Last fiddled with by Alberico Lepore on 20171226 at 22:31 Reason: add 3°example 

20171226, 22:59  #36 
Aug 2006
175B_{16} Posts 

20171226, 23:18  #37  
May 2017
ITALY
2×3^{2}×29 Posts 
Quote:
[22862368129858255125743027343921582130814586194768701417645652*a^2]=a*(an)=Z , Z2*(an)^2=an , a^2+n*a=2286236812985825512574302734392158213081458619476870141764565 

20171226, 23:31  #38 
Aug 2006
13533_{8} Posts 

20171226, 23:35  #39  
May 2017
ITALY
1012_{8} Posts 
Quote:
that is, you are unlucky if you catch a low number however it should be implemented and tested are you willing to do it? 

20171227, 03:50  #40  
Aug 2006
175B_{16} Posts 
Quote:
I don't see any likelihood of it succeeding, and my finances preclude me from taking on nonpaying freelance work at this time. But if you manage, please let us know. 

20171227, 09:44  #41 
May 2017
ITALY
2×3^{2}×29 Posts 
ENG
I have come to a conclusion (conjecture): when the difference in the product of the child nodes is equal to the difference of the product of the parent node then it means that a> (an) when the difference in the product of the child nodes is different from the difference in the product of the parent node then it means that a <(an) ITA Sono giunto ad una conclusione (congettura) : quando la differenza del prodotto dei nodi figli è uguale alla differenza del prodotto del nodo padre allora significa che a>(an) quando la differenza del prodotto dei nodi figli è diversa dalla differenza del prodotto del nodo padre allora significa che a<(an) Example 9967*6781=67586227 9967*6781=67586227  9967*6781=3186 6781*3595=24377695  67813595=3186 3595*409=1470355  3595409=3186 2777*409=1135793  2777409=2368  14287665=2077*5145  51452077=3068 1959*409=801231  ...... 1141*409=466669  ...... 409*323=132107  409323=86  202455=409*495  495409=86 323*237=76551  323237=86 237*151=35787  237151=86 151*65=9815  15165=86 65*21=483  6521=44 23*21=1365 FINE 2321=2 what do you think about it? Last fiddled with by Alberico Lepore on 20171227 at 10:05 Reason: Format 
20171227, 09:59  #42 
May 2017
ITALY
2×3^{2}×29 Posts 
ENG
If the conjecture was true then to face the problem of the small number (in this case the 409) we can make a log_2 at each step so a disadvantage would become an advantage ITA Se la congettura fosse vera allora per affrontare il problema del numero piccolo (in questo caso il 409) possiamo fare un log_2 ad ogni step così uno svantaggio diventerebbe un vantaggio what do you think about it? 
20171227, 14:41  #43 
May 2017
ITALY
2·3^{2}·29 Posts 
If the conjecture was true
the computational complexity is k * log_2 (N) where k is a constant that depends on the number that is, it is the depth of the tree without repetitions 
20171228, 00:10  #44 
Aug 2006
3×1,993 Posts 
If it depends on the number, it's not a constant. What kind of dependency does it have?

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