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

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• ... 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
> 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.

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...
• ... 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