CJam, 356 186 168 131 126 111 99 96 94

"_|`'~,Y/G>z`|"_~

Try it online in the CJam interpreter.

Output

"_|`'~,Y/G>z[ \$&(*.02468:<@BDFHJLNPRTVXZ^bdfhjlnprtvx!#%)+-13579;=?ACEIKMOQSUW]acegikmoqsuwy{}

Idea

Using a variation of a technique common in CJam quines, we sort the printable ASCII characters by whether they appear in the source code, and the non-appearing ones – with one exception – by their code points' parity bits.

With the proper source layout, we also manage to sort the source code characters – with one exception – by their frequencies.

Special care has to be taken that two adjacent characters do not appear one after the other for the first time in the source code, as this would invalidate the answer.

Code

"             "_~  Push a string, duplicate it and evaluate the copy.
 _|                Perform the set union of the original string with itself.
                   This is just an "excuse" to introduce the underscore.
   `               Inspect the string (surrounds it with double quotes).
    '~,            Push the string of Unicode characters before the tilde.
       Y/          Divide it into pairs.
         G>        Discard the first 16 pairs (control characters).
           z       Zip. This interleaves the pairs, ordering the characters
                   by their code points' parities.
            `      Inspect the array, i.e., push its string representation.
             |     Perform set union with the string of source code characters.

Brainfuck, 1692 826 765

(Still) Unoptimized, I know. I'm working on it (leave opts in the comments).

++[------>+<]>.++.---[-->+++<]>-.[->+++++++++<]>.--[----->+<]>.[-->+++<]>+.++.>+++[<---------->-]<-[-->+<]>-++.>++[>+++<-]>[<<++.>>-]<<++++.++.++.++.++.++.++++>>++++[<++++++++>-]<[<++.>-]<---.++.-->+++[>+++++<-]>-[<<--.>>-]<<---->++++[>+++++<-]>++[<<--.>>-]<<------.--.--.--.--.

Output:

+->.<[] "$&(*,02468:@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~{}ywusqomkigeca_YWUSQOMKIGECA?=;97531/)'%#!

I'm already utilizing overflow on 8bit cells to some extent, but I guess you could still optimize it. Though that would decrease the use of cheap chars :).

Pyth, 173 170

Code

-so%CN2rd\~p"p~\dr2NC%os-

Output

p~\dr2NC%os- "$&(*,.0468:<>@BDFHJLPRTVXZ^`bfhjlntvxz|!#')+/13579;=?AEGIKMOQSUWY[]_acegikmquwy{}

Hardcoding a quine-like string. Conveniently, the " character is very near the start of the generated string. Prints even then odd characters after the "quine".

Much thanks to Dennis for saving 3 points, and making the code a palindome!

Try it here

Java, 3518 3189 2692

A simple loop that prints even characters, then odds. I tried a few things to optimize earlier ASCIIs, but most ended up making it longer overall, and ended up with a higher score.

void A(){for(char A=31;A!=126;System.out.print(A+=2))A=A==125?30:A;}

Output is:

!#%')+-/13579;=?ACEGIKMOQSUWY[]_acegikmoqsuwy{} "$&(*,.02468:<>@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~

Edit: Misunderstood the scoring at first. After flipping it to odd first, then even, it scores a lot better.

C,42 bytes, score 1539

main(i){for(;i-191;i+=2)putchar(32+i%95);}

!#%')+-/13579;=?ACEGIKMOQSUWY[]_acegikmoqsuwy{} "$&(*,.02468:<>@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~

C,39 bytes, score 1687

main(i){for(;i-96;)i=putchar(32+i%95);}

!Aa"Bb#Cc$Dd%Ee&Ff'Gg(Hh)Ii*Jj+Kk,Ll-Mm.Nn/Oo0Pp1Qq2Rr3Ss4Tt5Uu6Vv7Ww8Xx9Yy:Zz;[{<\|=]}>^~?_ @`

In both cases, i is initialised to the number of strings on the commandline (as no arguments are given, this is 1.)

The first version does things the obvious way, incrementing by 2, taking modulo 95 and thefore printing all the odds then all the evens.

The second version takes advantage of the fact that putchar returns the character printed. As 32 is coprime to 95, we can cycle through the characters. As C contains a lot of lowercase characters I hoped that this, besides being shorter, would have a lower score but unfortunately this is not the case.

Befunge-93, 801 797

Code:

84*::         ^
v+!0  _@#$    <
>:,2+:882**2-`|
^             <

Output:

 "$&(*,.02468:<>@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~!#%')+-/13579;=?ACEGIKMOQSUWY[]_acegikmoqsuwy{}

You can try it here if you want.

It works by outputting 32-126 evens, and then 33-125 odds. If anyone wants an explanation, I'd be willing to.

I golfed it until I got it better than brainf***, which I deemed to be the lowest I could go. As far as golfing strategies, I generated the ascii characters and then tried to replace costly characters with cheaper ones (like 1 with 2). I found out since g was so expensive, it was better to compute 126 every iteration. I also wrapped around the top since ^ was cheaper than v.

801 -> 797: Recent change was removing extra spaces that was a relic from using g.

I might come back to it later to improve, but I'm satisfied with it for my first golfing attempt.

Octave, 628

Code

["" 32:2:126 33:2:125]

Output:

 "$&(*,.02468:<>@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~!#%')+-/13579;=?ACEGIKMOQSUWY[]_acegikmoqsuwy{}

Two ranges implicitly converted to string. Not sure if returning as Ans is acceptable, also gives a warning about the implicit conversion. Tried some other range vectors, but could not find anything more efficient.

Ruby 2.2, 1157

eval s='srand 1;([*s.bytes].shuffle|[*33..0x7e].shuffle).map{|c|putc c}'

Output:

f.p|cahu]xens7*0{)3tbmdy[}l1; r(o@&gN/MjzSVv~>D4I`L\KB92=i%PHE?5TQw,W-#6U'^Y!$R"XkO_q+CAGZF<8:J

This is a pretty dumb solution (and I'm not sure srand shouldn't be a standard loophole especially since it drastically reduces portability). Shuffles (most of) the bytes in its own source code and shuffles the rest, then uniques and concatenates the arrays. Uses a random seed picked so that the output is legal (the fact that it's a single digit is pure luck).

Haskell, 1660 1376

""!_="O"
(a:b)!(c:d)=a:c:b!d
a=[' '..'N']!['P'..]

Defines the function a which returns the string:

 P!Q"R#S$T%U&V'W(X)Y*Z+[,\-].^/_0`1a2b3c4d5e6f7g8h9i:j;k<l=m>n?o@pAqBrCsDtEuFvGwHxIyJzK{L|M}N~O

CBM BASIC V2, 2553

1FORI=0TO47:PRINTCHR$(32+I*2);:NEXT
2FORI=0TO47:PRINTCHR$(33+I*2);:NEXT 

the output (converted in ASCII by a python script on pc):

<blank>"$&(*,.02468:<>@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~!#%')+-/13579;=?ACEGIKMOQSUWY[]_acegikmoqsuwy{}

Haskell, 830

['!','#'..'}']++[' ','\"'..'~']

Evaluates to the string:

!#%')+-/13579;=?ACEGIKMOQSUWY[]_acegikmoqsuwy{} "$&(*,.02468:<>@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~

Inspired by @Jørgen's answer and completely different from my own.

Brainfuck, score 576 667

Thinking about it, 576 seemed to good to be true: I did a little estimation and worked out my score to be around 95*6 + 45*2 = 660. Something must have gone wrong the first time I ran the validator. The correct score is closer to my estimate. It's still not a bad score.

+++++++++++++++++++++++++++++++++++++++++++++.--.+++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.++.---.++.----.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.--.---.--.--.--.--.--.--.+++.--.++++.++.++.

Keep it simple.

Basically just walks up and down the ASCII set, printing characters. The three characters used in the program are printed first. Turning round at either end was a little bit tricky.

-+.02468:<>@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~{}ywusqomkigeca_][YWUSQOMKIGECA?=;97531/,*(&$" #!%')

BBC BASIC, 2554

Code

n=32
s$=""
REPEAT
  s$+=CHR$(n)
  n+=2
  IFn=128THENn=33
UNTILn=127
PRINTs$

Output

 "$&(*,.02468:<>@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~!#%')+-/13579;=?ACEGIKMOQSUWY[]_acegikmoqsuwy{}

Python 2, 72 bytes (3188) 116 bytes (1383) (1306)

thanks @FryAmTheEggman for the join trick ;)

thanks @nim (Did I misread the text? :P)

n=' nri(a)2:]o[c=fh1+t"3egj,p.7'
a=[chr(r)for r in range(32,127)if not chr(r)in n]
print n+"".join(a[::2]+a[1::2])

output:

 <blank>nri(a)2:]o[c=fh1+t"3egj,p.7!$&*/469<?ACEGIKMOQSUWY\_bkmsvxz|~#%'-058;>@BDFHJLNPRTVXZ^`dlquwy{}

Fortran 90, 1523 1519 1171

This is a nested output loop, similar to other answers. Not too confident that much improvement is possible...

PRINT*,((CHAR(J),J=L,126,2),L=32,33)
END

Output:

 "$&(*,.02468:<>@BDFHJLNPRTVXZ\^`bdfhjlnprtvxz|~!#%')+-/13579;=?ACEGIKMOQSUWY[]_acegikmoqsuwy{}

Edit: Forgot that Fortran 90 is necessary for this code, 77 requires code to start in the 7th column. On the other hand, the language is case insensitive, allowing an easy improvement. The loop counters are J and L because these are the first two letters in the output string implicitly declared as integers by Fortran.

JavaScript, 3791 3172 3169 2548 2144 2104 2071 1885 1876

Code

t=''
S=i=95
while(i--)t+=String.fromCharCode(i*11%S-
-32)
alert(t)

Output

ti^SH=2'{peZOD9.#wlaVK@5*~sh]RG<1&zodYNC8-"vk`UJ?4)}rg\QF;0%yncXMB7,!uj_TI>3(|qf[PE:/$xmbWLA6+ 

gawk, 2782 1988 1821

END{for(rrf=rrr="rf(3)+=;1\"$?:~ptoin[<-EN% ^.|P";fr++<333;$fr=(ff=sprintf("%c",fr))~"[[(]"?f:ff);for(;r++<33+13+1;rrf=f)printf(rrf)(rrr~(rr=$(31+1+r+r))?f:rr)(rrr~(rr=$(133-(3+3+r+r)))?f:rr)}

Output

rf(3)+=;1"$?:~ptoin[<-EN% ^.|P}{&yw*u,sq02m4k68gec>a@_B]DFYHWJULSQORMTKVIXGZ\CA`bd9h7j5l/vx'z#!

Usage

Copy and paste the following to your console
(mawk won't work, because it's too sctrict stricter with printf)

awk 'END{for(rrf=rrr="rf(3)+=;1\"$?:~ptoin[<-EN% ^.|P";fr++<333;$fr=(ff=sprintf("%c",fr))~"[[(]"?f:ff);for(;r++<33+13+1;rrf=f)printf(rrf)(rrr~(rr=$(31+1+r+r))?f:rr)(rrr~(rr=$(133-(3+3+r+r)))?f:rr)}' < /dev/null

The < /dev/null at the end signals the end of input, so the END block will be executed.

I basically interweaved the characters coming from the bottom and coming from the top. Then I analysed, which characters were used the most in the program and printed them first, in order of frequency. Then I had to make sure that no character is printed more than one time. The weaving in opposite directions made it more probable that an already used character wouldn't lead to printing neighbours. But they met in the middle at P, so I had to print that in the beginning too. Then there were some problems with characters which are used in regexps... Then I renamed the variables cheaply and did the whole thing over again. Then I found some characters I could replace in my program, and did the whole thing over again. And so on.. I finally tweaked the string with the preferred characters a little by testing.

I think I'm done :D

During the process I never executed the program from the command line, but constructed a string I executed from inside a script, which would analyse the output for correctness and give me the score and stuff. That score output helped a lot. Of course I rechecked here (you never know) but it got me the same result.

There the program looks like this

p=sprintf("END{"\
"for(rrf=rrr=%c%s%c;fr++<333;$fr=(ff=sprintf(%c%cc%c,fr))~%c[[(]%c?f:ff);"\
"for(;r++<33+13+1;rrf=f)printf"\
"(rrf)(rrr~(rr=$(31+1+r+r))?f:rr)(rrr~(rr=$(133-(3+3+r+r)))?f:rr)}"\
,34,s=sprintf("rf(3)+=;1%c%c$?:~ptoin[<-EN%c ^.|P",92,34,37),34,34,37,34,34,34)

Matlab, 763

Of course, it's quite impossible to beat the Octave solution in MATLAB, since it doesn't have " which is 'early' in the ASCII-range. However, I decided to get a bit creative and figured to abuse randperm. I admit that it's a bit hacky and some could consider it cheating, but I guess it's a nice touch. First, the program and output:

rng(1194663);['' randperm(95)+31]

Ouput:

p2)[]913r~jZe:'Xf +b(Atd@LHT*7&xmN>6!?CJgwsaSh|/McO4_EkK=$5VP-%D<"Gz#Yq08n};WB`{.l\Quy^vR,IFoiU

For calculating an appropriate seed, I used the following program, which I ran until seed=4648029 (i.e., until the dishes were done)

minscore=1e9;
bestseed=0;
for(seed=1:1e9)
    rng(seed)
    p=randperm(95)+31;
    if(any(abs(diff(p))==1))
        continue
    end
    codestring=sprintf('rng(%d);['''' randperm(95)+31]',seed);
    score=0;
    for(i=1:length(codestring))
        score=score + find(codestring(i)==p,1);
    end
    if(score<minscore)
        minscore=score;
        bestseed=seed;
    end
end