Cops and Robbers bowling (Cops) Cops' Thread
In this challenge, your goal is to write a function that accepts a integer N, and that function will return the first N primes.  However, you must do this in the most characters possible.

You may use any method you wish to produce the primes.
You are not allowed to modify your code after posting.
Only printable ASCII characters are allowed.
A character cannot appear in your code more than 100 times.

However, after posting your answer, the robbers will have the opportunity to crack your code by posting a modified version of your code.  Your code is considered cracked if:

The modified code has a maximum levenshtein-distance of 10 from yours, and has fewer characters.
The function, given the same valid input, will produce the same output in the same language of the same version
The robber posted the modified version within 168 hours (1 week) of your post.

If a robber failed to crack your code within 168 hours, your code is considered safe.  The winner of this challenge is the cop with the longest safe code.
The title of your thread should contain the language, the number of characters, and the current status (Open, Cracked, Safe) of your code.
The accompanying robbers thread can be found here
Relevant tools

Frequency counter
Character counter
Levenshtein distance

You can also use this snippet to calculate the Levenshtein distance:


var f=document.getElementById("f"),g=document.getElementById("s"); function h(){var a=f.value,e=g.value;if(128&lt;a.length)a="&lt;span style='color:red'&gt;First program is too long!&lt;/span&gt;";else if(128&lt;e.length)a="&lt;span style='color:red'&gt;Second program is too long!&lt;/span&gt;";else{if(0===a.length)a=e.length;else if(0===e.length)a=a.length;else{var d=[],b;for(b=0;b&lt;=e.length;b++)d[b]=[b];var c;for(c=0;c&lt;=a.length;c++)d[0][c]=c;for(b=1;b&lt;=e.length;b++)for(c=1;c&lt;=a.length;c++)d[b][c]=e.charAt(b-1)===a.charAt(c-1)?d[b-1][c-1]:Math.min(d[b-1][c-1]+1,Math.min(d[b][c-1]+1,d[b-1][c]+ 1));a=d[e.length][a.length]}a="Distance = &lt;strong&gt;"+a+"&lt;/strong&gt;"}document.getElementById("d").innerHTML=a}f.onkeyup=h;g.onkeyup=h;
&lt;h3 id=ft&gt;First program&lt;/h3&gt;
&lt;textarea id=f rows=7 cols=80 style="width:100%"&gt;&lt;/textarea&gt;
&lt;h3 id=st&gt;Second program&lt;/h3&gt;
&lt;textarea id=s rows=7 cols=80 style="width:100%"&gt;&lt;/textarea&gt;
&lt;p id=d&gt;&lt;/p&gt;



Credits to Doorknob!
Python, 9304 (Safe)

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.CE*vh|ZteK3L3N6**{sUtjV= 5Tefx^jS330yA/]%E[[y};I~..k{-&lt;fe~qIQ,o._`k~u+.(gjvw%D&lt;_NO'Vkk+bRlU1VPNNlH'Y&gt;TaUZNu!u[B7rg&lt;!4B%71A`pd#R~rH}VcT&gt;St0I`(-sqiAX#GUr!?i=|1Y`27A:iwE:Gr|K0EQ8V.isY`jRxfv1rVOe7H2hM)DUV&amp;qC`~t*xfe0ogqpLVFULG!cd^tDOwwGb_=DdDRp~Mm pRf+RO(Rjss&gt;}8].=t.'caJ2e+}hh_Z&gt;o22sqJotXJwW8U8k0fbf3z+p+Q+R":eval(eval(`s[:-28:11]+s[-28:]`+".decode('base64').decode('zip')"*10))


Similar to my previous answer, the long string is created through repeated application of zlib and base64 encoding. A short answer might be safer, but finding a longer one is more interesting.

To protect against string multiplication attacks and string concatenation attacks,
the following tricks are used:  


The extra characters not used in base64 encoding (added for more characters) are shuffled, and then interleaved in evenly spaced intervals.  
After the string is fully decoded and evaluated, there is still one last check to count the occurrences of each character in the string. 


Hopefully it won't get golfed by some unexpected syntax shortening attack.