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Re: [PrimeNumbers] Re: Prime chains x-->Ax+B [puzzle 11]

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  • Phil Carmody
    ... And you were only looking for an 8 by the looks of it - term 9 has 17 as a factor. ... Ditto, term 12 having a 23 as a factor. ... Corroborated: (and here,
    Message 1 of 143 , Dec 4, 2010
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      -- On Fri, 12/3/10, djbroadhurst <d.broadhurst@...> wrote:
      > --- In primenumbers@yahoogroups.com,
      > Kevin Acres <research@...> wrote:>
      > > Seems very difficult to get a 8/12
      >
      > I found it rather easy to get a 8/12:
      >
      > p=7;for(k=1,8,print(factor(p));p=96328365*p-674298544);
      > Mat([7, 1])
      > Mat([11, 1])
      > Mat([385313471, 1])
      > Mat([37116615999606371, 1])
      > Mat([3575382933574921687714871, 1])
      > Mat([344410792240175811180613435317371, 1])
      > Mat([33176528504850823203577211921154930229871, 1])
      > Mat([3195840747248174368104654975623363340731873292371,
      > 1])

      And you were only looking for an 8 by the looks of it - term 9 has 17 as a factor.

      > and not much harder to get a 9/12:
      >
      > p=3;for(k=1,9,print(factor(p));p=237862725*p-713588164);
      > Mat([3, 1])
      > Mat([11, 1])
      > Mat([1902901811, 1])
      > Mat([452629409458306811, 1])
      > Mat([107663664748893631236931811, 1])
      > Mat([25609172680658279861161720490056811, 1])
      > Mat([6091467598816933241588548501453247755681811, 1])
      > Mat([1448933082303802516987335475350335971266609083806811,
      > 1])
      > Mat([344647171299431744552466406656001243790997338184040724431811,
      > 1])

      Ditto, term 12 having a 23 as a factor.

      > but then it did indeed start to get tough.

      Corroborated: (and here, as 2 are for free, 7=9)

      v=vector(10);p1=2;B=11-x*p1;p=p1;for(k=1,12,print(p);if(k>2,v[k-2]=p);p=p*x+B)
      while(n=input(),r=0;for(k=1,10,if(ispseudoprime(subst(v[k],x,n)),r++,break));if(r>6,print(r," ",n)))

      7 29709057996454
      7 7263780549444
      7 151166246100
      7 65068011125074
      7 84114269577064
      7 76656931587534
      7 59402114114520
      7 119930340985104
      7 101318539628098
      7 116698909426848
      7 88188386883268
      7 149758381779964
      7 130740466545294
      7 156653024729950
      7 128889269017180
      7 212346115784100
      7 193981060196650
      7 183192393790770
      7 245317914267790
      7 253470779612920
      7 252069956363974
      7 281241306114990
      7 333285815039134
      7 306117971946744
      7 376527527757800
      7 347395912514014
      7 374726694989208

      7, 7, 7 - is all I'm gonna get limited to 7!??!!??!!?

      7 383891136345938
      7 389994524189634
      7 391059035903064
      7 449305599037824
      7 493191506880678
      8 491550335809204
      9 519815699116440

      Phew, that's a little better!

      7 595324234851644
      7 629916755202098
      7 655175951516384
      7 704886845640968
      7 706950011329310
      7 733442990291760
      7 807366280229158
      7 774858123263324
      7 778469035284238
      7 842319483923908
      7 874455217723220
      7 898020387684784
      7 888526524582104
      7 903179551644778
      7 973402495542698
      7 955060423888748
      7 1068303740684214
      7 1045881643505370
      7 1086729527030588
      7 1116917783358818
      9 1188368874866394
      7 1177489195413664
      7 1172652104044378
      7 1223056406479128
      7 1240318813279338
      7 1212277454482458
      7 1224158658202514
      7 1221132485837100
      7 1236501442911098

      Giving up on 2, 11. Will try 3, 11...
    • djbroadhurst
      ... Suppose that we want to start with a square and get a square. Then we must solve the Diophantine equation y^2 = a*x^2 + b For any pair (a,b), Dario will
      Message 143 of 143 , Jan 7, 2011
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        --- In primenumbers@yahoogroups.com,
        Kevin Acres <research@...> wrote:

        > x=a*x+b either is a square or has a square as a major factor.

        Suppose that we want to start with a square and get a square.
        Then we must solve the Diophantine equation
        y^2 = a*x^2 + b

        For any pair (a,b), Dario will tell us all the solutions:
        http://www.alpertron.com.ar/QUAD.HTM

        David
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