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Another large pair of consecutive smooth numbers

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  • WarrenS
    Let f(x) = x*(x-11)*(x-24)*(x-65)*(x-90)*(x-129)*(x-173)* (x-212)*(x-237)*(x-278)*(x-291)*(x-302)/67440294559676054016000 then the smooth pair is Q and Q+1
    Message 1 of 2 , Mar 20, 2012
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      Let
      f(x) = x*(x-11)*(x-24)*(x-65)*(x-90)*(x-129)*(x-173)*
      (x-212)*(x-237)*(x-278)*(x-291)*(x-302)/67440294559676054016000
      then the smooth pair is
      Q and Q+1 where

      Q=f(2037335926565)
      =
      7582846503419980985727018264244718835660110144701285792357\
      0157185560815861224078102966540735794780952727297976931615\
      819922375
      =
      5^3*11*13*23*29*37*53*59*73*101^2*109*151*193*277*
      293*563*617*743*821*947*1459*2097803*1220327*718139*
      97523*54581*28807*207293*1610639*835633*80777*322939*11681*
      1847*4231*1871*2791
      =
      7.58 * 10^124 roughly

      and

      Q+1=
      2^3*3^5*7*41*43*47*67*89*113^2*223*431*433*461*491*577*1163*
      1249*1303*1543*337013*2573827*422069*123427*
      303587*36683*755173*3167*1682999*1914811*3593*17807*61031*
      2741*225149*5953

      max prime involved in either is 2573827=prime[188076].

      This one should be fairly difficult to beat, I think, even though I estimate
      that with this same prime bound an example with Q over 2000 digits long
      should exist (which I am not smart enough to find), albeit I think no Q over
      3000 digits long should exist.

      Although this might be beatable using the same idea and a larger search (and that would be interesting), it would be more interesting to employ a different idea.

      Another fairly strong example arises from Q=f(6457935020528)=7.80*10^130 roughly
      with maxprime=5881217=prime[405194].
    • djbroadhurst
      ... Definition: For positive integer n let p be the largest prime divisor of n*(n+1) and S(n) = log(n)/log(p) be the figure of merit for the smoothness of the
      Message 2 of 2 , Apr 20, 2012
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        --- In primenumbers@yahoogroups.com,
        "WarrenS" <warren.wds@...> wrote:

        > max prime involved in either 2573827

        Definition: For positive integer n let p be the largest
        prime divisor of n*(n+1) and S(n) = log(n)/log(p) be
        the figure of merit for the smoothness of the
        consecutive integers n and n+1.

        Then Warren's 125-digit example had

        S(n) =~ log(7.5828465*10^124)/log(2573827) =~ 19.480

        Here is a 135-digit example with greater merit:

        {f(y)=
        (y^2-11^4)*(y^2-35^2)*(y^2-47^2)*
        (y^2-94^2)*(y^2-146^2)*(y^2-148^2)/
        67440294559676054016000 - 1;}

        n = f(12971885307194);

        {if(type(n)=="t_INT",F=factor(n*(n+1))[,1];
        p=F[#F];print("n has "#Str(n)" digits");
        print("max prime is "p);
        default(realprecision,5);
        print("S(n) =~ "log(n)/log(p)));}

        n has 135 digits
        max prime is 6244451
        S(n) =~ 19.797

        David
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