Décomposition en facteurs premier en maple

Posté par vincprof vincprof

> lissolut3:=proc(n)
> L:=[1..n];
> k:=1;
> for a from 2 to n do
>  if evalb((a mod 2)=1)=true then
>     if  evalb(isprime(a-2))=false then
>        for j from 1 to evalf(floor((a^2)/4)) do  
>             if  evalb(deriv(j)<>a)=true then    
>             L[k]:=a;
>              k:=k+1;
>             fi;  
>        od;
>     fi;
>  fi;
> od:
> RETURN(L[1])
>  end:
Warning, `L` is implicitly declared local to procedure `lissolut3`

Warning, `k` is implicitly declared local to procedure `lissolut3`

Warning, `a` is implicitly declared local to procedure `lissolut3`

Warning, `j` is implicitly declared local to procedure `lissolut3`

> lissolut3(30);
>
Error, (in lissolut3) out of bound assignment to a list

Posté par vincprof vincprof

en rajoutant une boucle ca donne la meme chose :


lissolut3:=proc(n)
> L := [seq(i,i=1..n)];
> k:=1;
> for i from 1 to n do
> L[i]:=0;
> od;
> for a from 2 to n do
>  if evalb((a mod 2)=1)=true then
>     if  evalb(isprime(a-2))=false then
>        for j from 1 to evalf(floor((a^2)/4)) do  
>             if  evalb(deriv(j)<>a)=true then    
>             L[k]:=a;
>              k:=k+1;
>             fi;  
>        od;
>     fi;
>  fi;
> od:
> RETURN(L[1])
>  end:Warning, `L` is implicitly declared local to procedure `lissolut3`

Warning, `k` is implicitly declared local to procedure `lissolut3`

Warning, `i` is implicitly declared local to procedure `lissolut3`

Warning, `a` is implicitly declared local to procedure `lissolut3`

Warning, `j` is implicitly declared local to procedure `lissolut3`

> lissolut3(30);
>
Error, (in lissolut3) out of bound assignment to a list

Posté par vincprof vincprof

c'est bon j'ai trouvé une solution
> derivk:=proc(a)
> #k:=1;
> #while (deriv(k)<>a and k<=(a^2)/4) do  
> #            k:=k+1;
> #            if (deriv(k)=a)=true then    
> #            RETURN(false);      
> #            fi;
> #           true;
> #       od;
> for j from 1 to floor((a^2)/4) do
>       if deriv(j)=a then RETURN(false); fi;
> true;
> od;
> end:
>
>
Warning, `j` is implicitly declared local to procedure `derivk`

> derivk(10);

                                false

> lissolut2:=proc(n)
> L := [];
> k:=3;
> for a from 2 to n do
>  if evalb((a mod 2)=1)=true then
>     if  evalb(isprime(a-2))=false then
>       if (derivk(a))=true then L:=append1([L,[a]]); print(a);
>      fi
>     fi;
>  fi;
> od;
> RETURN(L)
>  end:
Warning, `L` is implicitly declared local to procedure `lissolut2`

Warning, `k` is implicitly declared local to procedure `lissolut2`

Warning, `a` is implicitly declared local to procedure `lissolut2`

> lissolut2(1000);
>

                                  3


                                  11


                                  17


                                  23


                                  29

Warning, computation interrupted



mais il a mis plus de 5 minutes pour trouver ca...