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 Register FAQ Search Today's Posts Mark Forums Read 2015-12-10, 18:50 #1 Dubslow Basketry That Evening!   "Bunslow the Bold" Jun 2011 40>> from mfaliquot import aliquot as aq >>> res = aq.mutation_possible(aq.Factors('2^9 · 3^2'), 241413849295789277550538965117493054292130551238033452890627062088880427173192445000089149654782010178364256923601584820904417153263) >>> print(aq.test_tau_to_str(res, 'C132', '\n')) Assuming that C132 is made of 2 primes, then since it's 7 (mod 8), it's possible that tau(n)=3=1+2 via the following conditions: p1%8==3, p2%8==5. Assuming that C132 is made of 2 primes, then since it's 15 (mod 16), it's possible that tau(n)=4=1+3 via the following conditions: p1%16==7, p2%16==9. Assuming that C132 is made of 2 primes, then since it's 15 (mod 32), it's possible that tau(n)=5=1+4 via the following conditions: p1%32==1, p2%32==15. Assuming that C132 is made of 2 primes, then since it's 47 (mod 64), it's possible that tau(n)=6=1+5 via the following conditions: p1%64==31, p2%64==49. Assuming that C132 is made of 2 primes, then since it's 111 (mod 128), it's possible that tau(n)=7=1+6 via the following conditions: p1%128==63, p2%128==81. Assuming that C132 is made of 2 primes, then since it's 239 (mod 256), it's possible that tau(n)=8=1+7 via the following conditions: p1%256==127, p2%256==145. Assuming that C132 is made of 2 primes, then since it's 239 (mod 512), it's possible that tau(n)=9=1+8 via the following conditions: p1%512==17, p2%512==255. Assuming that C132 is made of 3 primes, then since it's 7 (mod 8), it's possible that tau(n)=4=1+1+2 via the following conditions: p1%8==1, p2%8==3, p3%8==5. Assuming that C132 is made of 3 primes, then since it's 15 (mod 16), it's possible that tau(n)=5=1+1+3 via the following conditions: p1%16==1, p2%16==7, p3%16==9; p1%16==7, p2%16==13, p3%16==13; p1%16==5, p2%16==5, p3%16==7. Assuming that C132 is made of 3 primes, then since it's 15 (mod 32), it's possible that tau(n)=6=1+1+4 via the following conditions: p1%32==9, p2%32==15, p3%32==25; p1%32==5, p2%32==13, p3%32==15; p1%32==1, p2%32==1, p3%32==15; p1%32==15, p2%32==21, p3%32==29; p1%32==15, p2%32==17, p3%32==17. Assuming that C132 is made of 3 primes, then since it's 15 (mod 16), it's possible that tau(n)=7=2+2+3 via the following conditions: p1%16==3, p2%16==3, p3%16==7; p1%16==7, p2%16==11, p3%16==11. Assuming that C132 is made of 3 primes, then since it's 47 (mod 64), it's possible that tau(n)=7=1+1+5 via the following conditions: p1%64==13, p2%64==31, p3%64==53; p1%64==29, p2%64==31, p3%64==37; p1%64==25, p2%64==25, p3%64==31; p1%64==21, p2%64==31, p3%64==45; p1%64==31, p2%64==57, p3%64==57; p1%64==9, p2%64==31, p3%64==41; p1%64==1, p2%64==31, p3%64==49; p1%64==5, p2%64==31, p3%64==61; p1%64==17, p2%64==31, p3%64==33. Assuming that C132 is made of 3 primes, then since it's 111 (mod 128), it's possible that tau(n)=8=1+1+6 via the following conditions: p1%128==63, p2%128==101, p3%128==125; p1%128==25, p2%128==63, p3%128==121; p1%128==21, p2%128==63, p3%128==77; p1%128==37, p2%128==61, p3%128==63; p1%128==1, p2%128==63, p3%128==81; p1%128==29, p2%128==63, p3%128==69; p1%128==63, p2%128==109, p3%128==117; p1%128==33, p2%128==49, p3%128==63; p1%128==41, p2%128==63, p3%128==105; p1%128==9, p2%128==9, p3%128==63; p1%128==63, p2%128==97, p3%128==113; p1%128==5, p2%128==63, p3%128==93; p1%128==57, p2%128==63, p3%128==89; p1%128==63, p2%128==73, p3%128==73; p1%128==13, p2%128==63, p3%128==85; p1%128==17, p2%128==63, p3%128==65; p1%128==45, p2%128==53, p3%128==63. Assuming that C132 is made of 3 primes, then since it's 15 (mod 32), it's possible that tau(n)=8=2+2+4 via the following conditions: p1%32==15, p2%32==19, p3%32==27; p1%32==3, p2%32==11, p3%32==15. Assuming that C132 is made of 3 primes, then since it's 239 (mod 256), it's possible that tau(n)=9=1+1+7 via the following conditions: p1%256==101, p2%256==127, p3%256==189; p1%256==33, p2%256==113, p3%256==127; p1%256==5, p2%256==29, p3%256==127; p1%256==37, p2%256==127, p3%256==253; p1%256==13, p2%256==127, p3%256==149; p1%256==1, p2%256==127, p3%256==145; p1%256==9, p2%256==73, p3%256==127; p1%256==127, p2%256==137, p3%256==201; p1%256==17, p2%256==127, p3%256==129; p1%256==121, p2%256==127, p3%256==217; p1%256==21, p2%256==127, p3%256==141; p1%256==127, p2%256==181, p3%256==237; p1%256==127, p2%256==161, p3%256==241; p1%256==41, p2%256==41, p3%256==127; p1%256==89, p2%256==127, p3%256==249; p1%256==127, p2%256==153, p3%256==185; p1%256==49, p2%256==97, p3%256==127; p1%256==45, p2%256==117, p3%256==127; p1%256==53, p2%256==109, p3%256==127; p1%256==93, p2%256==127, p3%256==197; p1%256==127, p2%256==173, p3%256==245; p1%256==127, p2%256==193, p3%256==209; p1%256==127, p2%256==177, p3%256==225; p1%256==65, p2%256==81, p3%256==127; p1%256==127, p2%256==133, p3%256==157; p1%256==25, p2%256==57, p3%256==127; p1%256==77, p2%256==85, p3%256==127; p1%256==105, p2%256==127, p3%256==233; p1%256==127, p2%256==205, p3%256==213; p1%256==61, p2%256==127, p3%256==229; p1%256==69, p2%256==127, p3%256==221; p1%256==125, p2%256==127, p3%256==165; p1%256==127, p2%256==169, p3%256==169. Assuming that C132 is made of 3 primes, then since it's 47 (mod 64), it's possible that tau(n)=9=2+2+5 via the following conditions: p1%64==27, p2%64==31, p3%64==35; p1%64==19, p2%64==31, p3%64==43; p1%64==11, p2%64==31, p3%64==51; p1%64==3, p2%64==31, p3%64==59. Better documentation is on my list of things to do, though Code: >>> from mfaliquot import aliquot as aq >>> help(aq)` should be a decent way to get started. Last fiddled with by Dubslow on 2015-12-10 at 18:53  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post wildrabbitt Software 7 2018-01-12 19:46 Matt Linux 1 2007-02-22 22:36 Istari Factoring 30 2005-07-12 20:20 Uncwilly Programming 9 2005-03-04 13:37 gbvalor Software 15 2004-01-04 11:51

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