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## Bug in Pari gp ?

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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] [8, 8] [9, 10] [10, 11] [11, 11] [12, 12] [13, 13] [14, 15] [15,
Message 1 of 2 , Jun 8, 2009
? 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]
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[24, 24]
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[26, 26]
[27, 28]
[28, 29]
[29, 30]
[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]
[8, 8]
[9, 9]
[10, 10]
[11, 10]
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[28, 28]
[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]
[7, 7]
[8, 8]
[9, 9]
[10, 9]
[11, 10]
[12, 11]
[13, 13]
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[29, 29]
[30, 30]
• ... 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
> 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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