Mathematica: Help with creating a more efficient algorithm

[Smidgen]

I've created a simple algorithm to count primes up to say 1000 which satisfies a certain criterion :

count = 0;
Do[p = Prime[jj];
If[And[MultiplicativeOrder[2, p] == p - 1,
MultiplicativeOrder[3, p] == p - 1], count = count + 1], {jj, 3, 1000}]
count

Now this algorithm works... but I'm interested in the percentage of primes (here, out of 1000) that satisfies the criterion. My algorithm only counts the primes but does not divide the number of primes counted by the number of primes. I would like to create a function where I can input any x ( the number of primes), and the function will spit out the percentage that I'm interested in. But I'm having a problem with this function ( particularly with Return[] ) that I'm testing:

func23[x_] :=
[
count = 0;
Do[p = Prime[jj];
If[And[MultiplicativeOrder[2, p] == p - 1,
MultiplicativeOrder[3, p] == p - 1], count = count + 1], {jj, 3,
x}];
Return[count/(x - 2)]
]

Any insights or suggestions would be greatly appreciated :)

 

[Dale]

The code you show has a syntax error. You need to replace the outermost [] with (). The Return is redundant, but not harmful.

 

[Smidgen]

DaleSpam, thanks for spotting it... I tried your suggestion and I was able to get the correct percentage I was looking for.

 

[Bill Simpson]

func23[x_] :=
Length[
Select[Range[3, Prime[x]],
PrimeQ[#] && MultiplicativeOrder[2, #] == #-1 && MultiplicativeOrder[3, #] == #-1 &
]
]/(x - 2)

OR

func23[x_] :=
Count[Range[3, Prime[x]], z_ /; PrimeQ[z] && MultiplicativeOrder[2, z] == z-1 && MultiplicativeOrder[3, z] == z-1]/(x - 2)

 

[Smidgen]

Bill Simpson, thanks for the additional codes... glad I could see different ways of arriving at the same percentage.