2021-09-23, 10:34 | #12 |
"Matthew Anderson"
Dec 2010
Oregon, USA
3^{2}×131 Posts |
more calculations
Hi again all,
I have probably done over 50 quick calculations for factordb.com. In My Humble Opinion, this is a useful database. Now we know that a C74 is a P22 times a P24 times a P29. Specifically, (2123766^17-1)/4830869344730522132314982799853985 <74> = 3995338558734555154151<22> · 483489332374281832645177<24> · 38980509326442350080033766609<29> this calculation took 135 seconds on my Intel i7, 2.93 GHz computer. Good fun Matt |
2022-02-22, 04:51 | #13 |
"Matthew Anderson"
Dec 2010
Oregon, USA
3^{2}·131 Posts |
Matt's factordb.com session tonight
I know you all are eager to see this.
For a laugh, see reddit.com/r/softwaregore This is code gore. # My comments were lost in translation, so you see "NULL". Code:
> NULL; > ifactor((2^303-26070)*(1/14318710184332826954770562)); %; Warning, computation interrupted > NULL; > NULL; > ifactor((2^303-12364)*(1/17111387030591077765288038823334466452)); (421924499807849134091) (2257193170479512364241080475925467) > NULL; > > ifactor((47^47-607868*3^151)*(1/563876793527)); (320053816119330053) (17749185239857015560248112454200033855665554126457) > NULL; Factordb.com wants these factorization s, and I have a tool that can do it, so I help out. Have a pleasant rest of your day. |
2022-02-22, 08:37 | #14 |
"Matthew Anderson"
Dec 2010
Oregon, USA
1179_{10} Posts |
More factordb.com
Hi again all,
Code:
> ifactor(12728090287417836575217719014652994601005209080166443917625601931139); (23) (71) (79) (97) (101) (103) (127) (131) (157) (251) (307) (311) (331) (449) (523) (661) (881) (937) (9511) (1831) (2371) (18691) (7561) (2731) (2311) > ifactor(83960497724513525996447551); (103) (137) (223) (271) (397) (433) (449) (853) (1153) (1297) > ifactor((47^47-607615*3^151)*(1/5888921458)); (3742690805057637823137877) (4695287137889797061) (30955787507078026189588333) > ifactor((2^303-13794)*(1/468401961602)); (270404512589470744637168966897) (1286637028865374347491025669817071695141439\ 61165631) > ifactor((2^303-28584)*(1/16450495022145781578813822166184)); (34855869486345119430765739853) (28420636125756948420304044911387) The above code represents 10 minutes of my computer's calculation time. Good fun. Matthew Last fiddled with by MattcAnderson on 2022-02-22 at 08:38 Reason: clean up code |
2022-02-22, 13:48 | #16 |
"Ed Hall"
Dec 2009
Adirondack Mtns
2^{2}·1,279 Posts |
Or, even PARI/GP (available in most linux repositories):
Code:
$ date;echo "factor((2^303-28584)*(1/16450495022145781578813822166184))" | gp -q;date Tue Feb 22 08:31:59 AM EST 2022 [ 34855869486345119430765739853 1] [28420636125756948420304044911387 1] Tue Feb 22 08:32:09 AM EST 2022 |
2022-02-22, 23:17 | #17 |
"Matthew Anderson"
Dec 2010
Oregon, USA
3^{2}×131 Posts |
I appreciate the suggestions, kar_bon and EdH.
For some reason, I prefer Maple, under a Windows platform. It is easier for me to use. At least you looked at my blog. You both get an A+. I am an ex school teacher, at community college level. Now retired. Regards, Matt Last fiddled with by MattcAnderson on 2022-05-30 at 02:23 Reason: spelling error |
2022-05-29, 10:28 | #18 |
"Matthew Anderson"
Dec 2010
Oregon, USA
3^{2}×131 Posts |
I did 3 factorizations tonight. I used Maple and it took 242 seconds. I know yafu is faster.
The composite numbers have 71 digits so 3 separate C71. > ifactor((10^77-902097)*(1/4617559)); (1231835793096446867189849001929203) (17580641889876672351849523974116725939) > ifactor((10^77-903192)*(1/2421224)); (44153726497395743090944141784629739) (935400648594425033776747820595443203) > ifactor((10^77-903567)*(1/7043737)); (18356243707875186787152019) (773415823472806954251581178472905383713563811) > Regards, Matt |
2022-05-30, 09:39 | #19 |
"Matthew Anderson"
Dec 2010
Oregon, USA
3^{2}×131 Posts |
I tried, unsuccessfully, to download yafu on my Windows computer. So I used Maple again.
850 seconds of computer time to do the 10 factorizations. So that's 14 minutes. plus the time it took me to cut and paste the numbers. > ifactor((10^79-71467)*(1/100516407)); (415196520508633322410371037) (239612427241508060164162583566211246413160687) > NULL; > ifactor((10^79-58711)*(1/169197177)); (32217712123101751) (3322354389083) (552161643662521620096524754679840484591029 ) > NULL; > ifactor((10^79-59127)*(1/134535283)); (10246441372319107150610777) (7254220359728226675028355218499318580974208203) > NULL; > ifactor((10^79-65147)*(1/746381351)); (786033472688938268962583638913) (17045046085811117653716230646110641196531) > NULL; > ifactor((10^79-88336)*(1/692835696)); (365851341700473666376338158849) (39451642843514306648014973762393822230441) > NULL; > ifactor((10^79-147445)*(1/277265805)); (92090012999657744473389857339) (391643708323158726317940739521054443982709) > NULL; > ifactor((10^79-193190)*(1/336498370)); (614639931266729357953) (48349976078148992975842315467923320911055809375821) > NULL; > ifactor((10^79-213027)*(1/247638251)); (455493389572979327) (191729565517439161) (462392845009718659471979432536148209 ) > NULL; > ifactor((10^79-217011)*(1/177863789)); (4171142998968471594400629914883503) (13478990935094332180998014639659382767) > NULL; > ifactor((10^79-294977)*(1/694650499)); (98291098940387286868762339) (146460144157043485287961013347211563879071143) > NULL; > All these composite numbers were 71 digits. I typed out the length of the prime factors, but it was lost in the cut and paste. You only see NULL; Have a nice day. |
2022-07-02, 00:20 | #20 |
"Matthew Anderson"
Dec 2010
Oregon, USA
3^{2}·131 Posts |
This is it.
Good times Look 58375189765560121426618297669679<32> Factor already known Found 2 factors and 0 ECM/P+1/P+1 results. Thank you for your contribution!! Result: status (?) digits number FF fully factored 74 digits 1104285714...57<74> = 741005168702867771<18> · 255288887770650551262373<24> · 58375189765560121426618297669679<32> New calculation - the story so far (star wars reference) 3964156213...71<74> factors as 76369348944813067995542418601892989907<38> times 1656496193897081<16> times 313358319590708066713<21> times 76369348944813067995542418601892989907<38>. Last fiddled with by MattcAnderson on 2022-07-08 at 05:03 |
2022-07-20, 17:14 | #21 |
"Matthew Anderson"
Dec 2010
Oregon, USA
3^{2}·131 Posts |
today's factordb session
Today I learned that
(10^79-253653)*(1/63408193995913273) <63> equals 1240407261075014928350683*127142374043023096518690934457531891833 in other words, we start with a Composite number (C) and find its prime factors. note that C63 means a positive composite whole number with 63 digits. a C63 equals a P25 times a P39. This is full prime factorization. A P39 is a 39 digit prime whole number, non-negative. I did the calculation that factordb.com asked for using Maple software. Got this message "Thank you for your contribution!!" similarly, (10^85-61574)/23838143056604878950854 equals 30957322240919310023385302747079353*(221309510703443*61229979274861) So this C63 equals a P35 * a P15 * a P14. Again, fully factored (FF) And a second "Thank you for your contribution!!" It makes me feel like I am doing something earth shattering :--) Now, to continue the exercise, (47^47-3^151*625186)/15013167471079231 and we have two nice little prime factors 10159422641338012069*20875181318585948571976024569935102678851739 So we see that C63=P20*P44. And our third "Thank you for your contribution!!" from factordb.com Having fun, I continue, (10^82+18525)/29418038266886705275 equals 1144138865405169632908059167*(34751873131457214461*8549275942075213) we have a C63 that is represented by 3 nice prime factors, A P28, a P20 and a P16. These calculations take about 20 seconds apiece on my rig. And of course, we see "Thank you for your contribution!!". Proceeding, (10^91+204520)/25623583168219463536366822120 equals 767341649510268970050796643*508594158619368915079248811240051147 This is summarized as a C63=P27*P36 and, of course, the kudos, "Thank you for your contribution!!" More, (10^79-338439)/57630465867129883 equals 10078773747059137326876880513*17216313947893011808535210545803259 Sumarized as C63=P29*P35 Last one for today, (10^77-188202)/388369050029614 equals 13224258579233485913*(10179633027616307963321489*1912722349862568901) So, another C63 is 'friable' as P20 * P26 * P19. I read about 'friable numbers' in an academic paper. According to Google, it means 'easily crumbled'. As in, the soil was friable between her fingers. At first, I thought it meant fryable, like can be cooked, like an egg is fryable. But the spelling is different. Composite numbers are friable into primes, with the right tool. 7 calculations with 119 seconds of computer 'calculating' time. Good fun. Have a nice day. Matt Last fiddled with by MattcAnderson on 2022-07-20 at 17:37 |
2022-11-04, 06:32 | #22 |
"Matthew Anderson"
Dec 2010
Oregon, USA
3^{2}·131 Posts |
more factordb.com exercises
Hi all, thought I would document some excercises for factordb.com tonight.
These are composite numbers without known factors, that is, until I find those factors. Seems like an unending quest for more integers. 1) 1023158077...47<65> = 42251827687945340008871<23> · 242157116825950976029865397595040631876557<42> that is 10231580773539729820091760424032485047135317471837448839756937147<65> now has its full prime number factorization. 23 seconds for Maple calculation 2) 2177776163...77<65> = 3183065695484865379709731<25> · 6841756883316582905918167565310671762767<40> so 21777761632132463921562492356618166667408889079359127924553385677<65> is a friable (composite) number. 3) 2203644330...59<65> = 318901290004477973928655822387<30> · 69101141942505843993250048862248057<35> similarly, 22036443306247652594045078394976506803452505721647364050927852059<65> has two prime factors. Have a nice day. |
Thread Tools | |
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Matt's various websites | MattcAnderson | MattcAnderson | 6 | 2022-04-17 05:01 |
Matt's 10 tuple thread | MattcAnderson | MattcAnderson | 7 | 2022-02-27 14:48 |
Matt's pairs procedure | MattcAnderson | MattcAnderson | 2 | 2022-02-22 23:14 |
Matt's 3 tuple thread | MattcAnderson | MattcAnderson | 1 | 2022-02-21 15:37 |
Matt's Sandbox | MattcAnderson | MattcAnderson | 20 | 2021-04-08 04:23 |