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Bug in Pari gp ?

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  • sopadeajo2001
    ? for(n=1,30,print([n,ceil(sqrtn(n^2,2))])) [1, 1] [2, 2] [3, 4] [4, 4] [5, 5] [6, 7] [7, 8] [8, 8] [9, 10] [10, 11] [11, 11] [12, 12] [13, 13] [14, 15] [15,
    Message 1 of 2 , Jun 8, 2009
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      ? for(n=1,30,print([n,ceil(sqrtn(n^2,2))]))
      [1, 1]
      [2, 2]
      [3, 4]
      [4, 4]
      [5, 5]
      [6, 7]
      [7, 8]
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      [30, 31]

      ? for(n=1,30,print([n,floor(sqrtn(n^2,2))]))
      [1, 1]
      [2, 2]
      [3, 3]
      [4, 4]
      [5, 5]
      [6, 6]
      [7, 7]
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      [29, 29]
      [30, 30]

      Is it a problem with the sqrtn(x,n) (n-th root of x)? Or wih floor and ceil functions?

      I wanted to search if some numbers were perfect powers using the trick:
      If(floor(sqrt(x,n))==ceil(sqrt(x,n)),.....

      and trying to use n (the n-th root ) as a variable in a for loop to look for many possible n-roots if this was possible technically.


      What is the way to solve my problem to detect if some kind of numbers are perfect powers?


      ? for(n=1,30,print([n,floor(sqrtn(n^3,3))]))
      [1, 1]
      [2, 2]
      [3, 3]
      [4, 4]
      [5, 5]
      [6, 6]
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      [28, 28]
      [29, 29]
      [30, 30]
    • Peter Kosinar
      ... Perhaps ispower() is what you re looking for? Peter -- [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
      Message 2 of 2 , Jun 9, 2009
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        > What is the way to solve my problem to detect if some kind of numbers are
        > perfect powers?

        Perhaps ispower() is what you're looking for?

        Peter

        --
        [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
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