PICO-8, 131 bytes

cls()function _draw()for i=1,9do for j=i-9,0do for k=-1,1,2do for l=0,1do print(i,72*l-(l-0.5)*(i-j)*8,48+k*j*6)end end end end end

Output:

Ungolfed version:

cls()                               //clear screen
function _draw()                    //keeps commandline off
    for i=1,9 do                    //numbers
        for j=i-9,0 do              //position control
            for k=-1,1,2 do         //y axis switch
                for l=0,1 do        //x axis switch
                    print(i,72*l-(l-0.5)*(i-j)*8,48+k*j*6)
                end
            end
        end
    end
end

FreeBASIC, 100 characters

Dim As Byte i,j,k
for i=-8 to 8 : for j=-8 to 9
k=abs(i)+abs(j)
if k > 8 then print " "; : else : ? chr(57-k);
next
? 
next 
Sleep
End

Vyxal C, 17 bytes

»÷rṡF»S¦øṁ½ƛ÷⋎;øm

Explanation

»÷rṡF»S                - 123456789 as string
       ¦               - Take all the prefixes
        øṁ             - Vertical mirror
          ½ƛ÷⋎;        - Halve and split, then join while removing the duplicate ends.
               øm      - Horizontal palindromize.
<C flag>               - Center and join list by newlines
<implicit output>      - Print

Try it Online!

jq -nr, 55 bytes

range(17)-8|fabs|" "*.+"123456789"[:9-.]+"87654321"[.:]

fabs is not supported on TIO but you can try it on https://jqplay.org/.

Alternatively try it on TIO with .*.|sqrt instead of fabs, for 59 bytes.

jq -rn, 76 65 60 59 bytes

-6 bytes thanks to ovs and another -6 (plus -5 more indirectly) thanks to Michael Chatiskatzi

def r:range(9),7-range(8);r|[8-.-r|"(" "*.//1-.)"[:1]]|add

Try it online!

Explanation

The -n flag specifies that there is no input; the -r flag outputs the result strings, each on its own line, without quotes.

def r:         Define a helper function r
range(9),      which generates the numbers 0 through 8, followed by
7 - range(8);  the numbers 7 through 0
               Main program:
r |            Start with the results of r
[              For each of those numbers, put the following in a list:
 8 - . - r |    8, minus the current number, minus the results of calling r again
 "(            For each of those numbers, interpolate the following into a string:
  " " * .        Try to repeat a space that many times
  //             and if the result is null (i.e. the number was less than 1),
  1 - .          subtract the number from 1 instead
 )"
 [:1]           Take the first character of each resulting string
] |            Take each resulting list of strings and
add            concatenate it together

To help visualize what's going on, here's what just the r|[8-.-r] part does:

[8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8]
[7,6,5,4,3,2,1,0,-1,0,1,2,3,4,5,6,7]
[6,5,4,3,2,1,0,-1,-2,-1,0,1,2,3,4,5,6]
[5,4,3,2,1,0,-1,-2,-3,-2,-1,0,1,2,3,4,5]
[4,3,2,1,0,-1,-2,-3,-4,-3,-2,-1,0,1,2,3,4]
[3,2,1,0,-1,-2,-3,-4,-5,-4,-3,-2,-1,0,1,2,3]
[2,1,0,-1,-2,-3,-4,-5,-6,-5,-4,-3,-2,-1,0,1,2]
[1,0,-1,-2,-3,-4,-5,-6,-7,-6,-5,-4,-3,-2,-1,0,1]
[0,-1,-2,-3,-4,-5,-6,-7,-8,-7,-6,-5,-4,-3,-2,-1,0]
[1,0,-1,-2,-3,-4,-5,-6,-7,-6,-5,-4,-3,-2,-1,0,1]
[2,1,0,-1,-2,-3,-4,-5,-6,-5,-4,-3,-2,-1,0,1,2]
[3,2,1,0,-1,-2,-3,-4,-5,-4,-3,-2,-1,0,1,2,3]
[4,3,2,1,0,-1,-2,-3,-4,-3,-2,-1,0,1,2,3,4]
[5,4,3,2,1,0,-1,-2,-3,-2,-1,0,1,2,3,4,5]
[6,5,4,3,2,1,0,-1,-2,-1,0,1,2,3,4,5,6]
[7,6,5,4,3,2,1,0,-1,0,1,2,3,4,5,6,7]
[8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8]

Any number that's greater than 0 gets turned into that many spaces (and then trimmed back to a single space). The other numbers gets subtracted from 1 (turning 0 .. -8 into 1 .. 9) and stringified.

Excel, 80 bytes

=CONCAT(LET(x,ABS(9-ROW(1:17)),y,9-ABS(9-COLUMN(A:R)),IFS(y>x,y-x,y," ",1,"
")))

Link to Spreadsheet

Regenerate, 109 bytes

(( {#2-1}| {8})($3${#3+1}|1)(${#4+1}$4|)n){9}(( $6| )(${$7/10}|${$3/10})(${$8-1+1/$8*#7-1/$8}|7){#7-1}n){8}

Try it here!

Explanation

The solution is divided into two parts, which generate the upper half of the diamond (including the center line) and the lower half, respectively. The upper half consists of group 1, repeated nine times (...n){9}; the lower half consists of group 5, repeated eight times (...n){8}.

The first part of each group handles the spaces. (I've replaced space with underscore in the explanations for better visibility.)

Upper-half spaces:

( {#2-1}| {8})
(            )  Group 2:
 _{#2-1}         Space, repeated one less time than the previous length of group 2
        |        Or, if this is the first row and there is no previous value of group 2...
         _{8}    Space, repeated eight times

Lower-half spaces:

( $6| )
(     )  Group 6:
 _$6      Space, concatenated to the previous value of group 6
    |     Or, if this is the first row and there is no previous value of group 6...
     _    A single space

Now for the diamond itself. It can be divided into four quadrants, and I used three different approaches to generate them:

Top left & top right: concatenation

We generate the left half (including the center) of each row with group 3, and the right half with group 4. For both halves, observe that we can generate the next row by concatenating a digit to the previous row:

 1234 321
   |   |
   v   v
12345 4321

So we can use a similar approach to what we did for the bottom-half spaces. The only difference is that the character we're concatenating at each step needs to be calculated.

Top left quadrant:

($3${#3+1}|1)
(           )  Group 3:
 $3             Previous value of group 3
   ${    }      concatenated to the result of...
     #3+1       Previous length of group 3 plus 1
          |     Or, if this is the first row and there is no previous value of group 3...
           1    The digit 1

Top right quadrant:

(${#4+1}$4|)
(          )  Group 4:
 ${    }       The result of...
   #4+1        Previous length of group 4 plus 1
        $4     concatenated to the previous value of group 4
          |    Or, if this is the first row and there is no previous value of group 4...
               Empty string

Bottom left: integer division

We generate the left half (including the center) of each row with group 7. Observe that each row is obtained by removing the rightmost digit from the previous row:

12345
  |
  v
 1234

If we treat the entire thing as an integer, we can perform this transformation easily by dividing it by ten:

(${$7/10}|${$3/10})
(                 )  Group 7:
 ${     }             The result of...
   $7/10              Previous value of group 7, divided by 10
         |            Or, if there is no previous value of group 7...
          ${     }    The result of...
            $3/10     Group 3 (the final upper-left quadrant row) divided by 10

Bottom right: digit-by-digit

There isn't an easy way to remove a digit from the left side of a previous match, so for the right half of each bottom row, we resort to generating one digit at a time with group 8. Here are the first three rows that we need to generate:

7654321
654321
54321

Each digit is one less than the previous digit, except that we have to "reset" when we hit 1. The reset value is always one less than the length of the left half of the row, which we have access to as group 7.

How can we tell when we've hit 1? I used an integer-division trick I first developed for programming in Acc!!: if N is a positive integer, then 1/N will be one iff N = 1 and zero if N > 1. Multiplying this quantity by (length of group 7) - 1 gives us the reset we need.

(${$8-1+1/$8*#7-1/$8}|7){#7-1}
(                      )        Group 8 (generates one digit at a time):
 ${                 }            The result of...
   $8-1                          Previous value of group 8, minus 1
       +     #7                  Plus the length of group 7
        1/$8*                    if the previous value of group 8 equals 1
               -1/$8             Minus 1 if the previous value of group 8 equals 1
                     |           Or, if there is no previous value of group 8...
                      7          The digit 7
                        {    }  Generate this many digits on each row:
                         #7-1   One less than the length of group 7

C (gcc), 82 78 bytes

#define W printf("%*lldn",i,1ll*j*j)
g(i,j){i<17&&g(W,j*10+1);W;}D(){g(9,1);}

Try it online!

Duocentehexaquinquagesimal, 58 bytes

(Úv×H.F°ǝ=ÉÈNŸ=~ʒ€
₁˨Dûl`N₆rÇX³ΣJd‘×∞*Åv€rØ(0՜֤ΔNā—St€ô

Try it online!

JavaScript (Node.js), 107 bytes

b=Math.abs,e="";for(i=0;i<17;i++){for(j=0;j<17;j++){z=9-(b(i-8)+b(j-8));e+=z>0?z:" "}e+="n"}console.log(e)

Try it online!

Templates Considered Harmful, 267 255 bytes

Ap<Fun<If<A<1>,Cat<Ap<Fun<Cat<Ap<Fun<If<A<1>,Cat<Ap<A<0>,Sub<A<1>,T>>,SP>,LF>>,Sub<I<9>,A<1>>>,Ap<Fun<If<lt<A<1>,A<2>>,Cat<Cat<A<1>,Ap<A<0>,Add<A<1>,T>,A<2>>>,A<1>>,A<1>>>,T,A<1>>>>,If<lt<A<1>,I<9>>,A<1>,Sub<I<18>,A<1>>>>,Ap<A<0>,Sub<A<1>,T>>>,LF>>,I<17>>

Try it online!

Has leading and trailing whitespace -- sorry! tch has no concept of "empty string" (at least not that I've found)

I doubt the ungolfed version would answer any questions (probably the opposite). I like to think of TCH as a lisp-like language with no functions, loops, or variables, so all the logic is done through recursive lambdas.

The basic structure:

lambda:  (run lower lambdas with int 1,2,3...,17,18, join output)
   lambda:  (convert input, run lower lambdas, join output)
     lambda:  (int -> string of spaces & newline)
     lambda:  (int -> string of numbers)

Haskell, (complete program) 79 bytes

main=mapM putStrLn[(' '<$[n..8])++(s n>>=show)|n<-s 9]
s n=[1..n-1]++[n,n-1..1]

Try it online!

Julia 1.0, 57 53 bytes

!i=[1:i;i-1:-1:1]
!9 .|>i->print.([' '^(9-i);!i;"n"])

Try it online!

Vyxal, C, 9 bytes

9ƛɾømĴ;ḆJ

Try it Online!

Lua, 80 bytes

for i=-8,8 do x=9-math.abs(i)n=math.ceil(10^x-1)//9print((" "):rep(9-x)..n*n)end

Try it online!

Zsh, ⁹¹ 61 bytes

eval s+={1..9}';<<<${(l:8:)s%?}`rev<<<$s`;'>f;{<f;tac f}|uniq

Try it online!

eval s+={1..9}';<<<${(l:8:)s%?}`rev<<<$s`;'>f;{<f;tac f}|uniq

eval    {1..9}'                          ;'                    # evaluate this 9 times
     s+=       ;                                               #  append the number to s
                <<<                                            #  print
                   ${      s  }                                #   s
                            %?                                 #    with the last character removed
                     (l:8:)                                    #    padded to 8 spaces
                               `rev<<<$s`                      #   then s, reversed
                                          >f;                  # all output to the file f
                                              <f;              # print f
                                                 tac f         # print f in reverse
                                             {        }|uniq   # remove the duplicated line in the middle
```

MathGolf, 11 bytes

9╒ñmÆ╒ñyFΩn

Try it online.

Explanation:

9╒           # Push a list in the range [1,9]
  ñ          # Palindromize it: [1,2,3,4,5,6,7,8,9,8,7,6,5,4,3,2,1]
   m         # Map over each inner integer,
    Æ        # using the following five character as inner code-block:
     ╒       #  Convert the integer to a list in the range [1,n]
      ñ      #  Palindromize it similar as before
       y     #  Join the list together to a string
         Ω   #  Prepend/append potential leading/trailing spaces (centralize)
        F    #  to make the string length 17
          n  # After the map: join the strings in the list by newlines
             # (after which the entire stack is output implicitly as result)

C (gcc), 171 163 147 142 bytes

-5 bytes thanks to @ceilingcat

#define f(x,y);for(j=0;j++<(i<9?x:y);)printf(
j;D(i){for(i=0;++i<19;puts("")){f(9-i,i-9)" ")f(i,18-i)"%d",j)f(i-1,17-i)"%d",i<9?i-j:18-i-j);}}

Try it online!

This code first prints a certain amount of spaces and then the numbers.

166 bytes

i,j;D(){for(;++i<19;){for(j=0;++j<19;i<10?(i+j>9&&j-i<9?printf("%d",j<9?i+j-9:9-j+i):printf(" ")):i-j<9&&i+j<27?printf("%d",j<9?j-i+9:27-i-j):printf(" "));puts("");}}

Try it online!

This code always chooses whether to print a space or a number, depending on the values of i and j.

Java 10 IntStream, 230 chars

interface A{static void main(String[]a){f(i->{var o=System.out;f(j->o.print(i+j>8?" ":""+(9-i-j)));o.println();});}static void f(java.util.function.IntConsumer c){java.util.stream.IntStream.range(-8,9).map(Math::abs).forEach(c);}}

Try it online!

Readable:

interface A {
    static void main(String[] a) {
        f(i -> {
            var o = System.out;
            f(j -> o.print(i + j > 8 ? " " : "" + (9 - i - j)));
            o.println();
        });
    }

    static void f(java.util.function.IntConsumer c) {
        java.util.stream.IntStream.range(-8, 9).map(Math::abs).forEach(c);
    }
}

Python 3, 54 bytes

saved another 3 bytes by letting the int function crash instead of a loop condition. Thanks to dingledooper.

i=8
while[print(f'{int("1"*(9-abs(i)))**2:^17}')]:i-=1

Try it online!

Python 3, 57 bytes

saved 1 byte by using f-Strings thanks to dingledooper.

i=8
while i+9:print(f'{int("1"*(9-abs(i)))**2:^17}');i-=1

Try it online!

Python 3, 58 bytes

Even shorter solution by a different approach (prints one extra space at the beginning of each line).

i=8
while i+9:j=abs(i);i-=1;print(' '*j,int('1'*(9-j))**2)

Try it online!

Python 3, 61 bytes (without extra spaces)

i=8
while i+9:j=9-abs(i);i-=1;print(f'%{8+j}d'%int('1'*j)**2)

Try it online!

These are my previous solutions:

Python 3, 66 bytes

f=lambda c:f'%{8+c}d'%int('1'*c)**2
for c in f(9):print(f(int(c)))

Try it online!

Python 3, 71 bytes:

f=lambda c:str(int('1'*int(c))**2)
for c in f(9):print(f(c).center(17))

Try it online!

Python 3, 123 bytes

k=[]
for t in range(10):
    l=str(111111111**2)[:t]
    k.append(l.rjust(9)+l[:-1][::-1])
print("n".join(k+k[:-1][::-1]))

Try it online!

JavaScript (Node.js), 73 bytes

for(z of(s=z=>"".padEnd(9-z)+(BigInt(10**z)/9n)**2n)(9))console.log(s(z))

Try it online!

Clojure, 115 bytes

(defn s[n](concat(range 1 n)(range n 0 -1)))(print(apply str(mapcat #(concat(repeat(- 9 %)" ")(s %)'("n"))(s 9))))

Try it online!

Shortened from 127 to 121 to 115.

T-SQL, 104 bytes

Inspired by @BradC's answer

DECLARE @ INT=8a:PRINT
space(abs(@))+stuff('12345678987654321',9-abs(@),abs(@)*2,'')SET
@-=1IF~@<8GOTO a

Try it online

Clojure, 160 bytes

(run! #(println(reduce str "" %))(map (fn[i](map #(let[x(- (if(> % 9)(- 9(- % 9))%)(- 9 i))](if(pos? x)x " "))(range 1 18)))(concat(range 1 10)(range 8 0 -1))))

JavaScript (V8), 75 bytes

for(a=9;b=--a>0?9-a:9+a;print(s))for(s='',c=9;--c+b;)s+=c<0?b+c:c<b?b-c:' '

Try it online!

JavaScript (V8), 75 bytes

for(b of x="12345678987654321")print(b-9?"".padEnd(9-b)+"1".repeat(b)**2:x)

Try it online!

Doing the calculations 1**2, 11**2, 111**2, ...

This solution fixes the precision problem. It correctly produces a 1 at the end of the longest line instead of 0. I think, this one is the clearest of the short solutions for JavaScript.

Instead of x="12345678987654321" you could also write x=111111111**2/10+"1" which is equally short.

Javascript, 103 96 93 89 88 87

f=(i=[...Array(17)].map((x,i)=>i<8?i:i-2*(i-8)))=>i.map(x=>i.map(y=>x+y>7?(x+y-7):' ').join``).join`n`

I haven't looked at the other answers up until this point, and just started out with this spreadsheet. Just saw that this answer isn't too bad of a golf compared to the other Javascript answers.

I tried to be clever coming up with something that generates a particular array (i) that I use for the x and y-axis. But it turns out, spelling out that array is shorter.

f=(i=[0,1,2,3,4,5,6,7,8,7,6,5,4,3,2,1,0])=>i.map(x=>i.map(y=>x+y>7?(x+y-7):' ').join``).join`n`

Update. Spelling out the array differently, but I'm sure that I can golf this further

f=(i=[...'01234567876543210'])=>i.map(x=>i.map(y=>1*x+1*y>7?(1*x+1*y-7):' ').join``).join`n`

Last update for tonight:

f=(i=[...'01234567876543210'])=>i.map(x=>i.map(y=>+x+1*y>7?+x+1*y-7:' ').join``).join`n`

Getting rid of one more byte:

f=(i=[...'0'+10*11111111**2])=>i.map(x=>i.map(y=>+x+1*y>7?+x+1*y-7:' ').join``).join`n`

And one more:

f=(i=[0,...11111111**2+'0'])=>i.map(x=>i.map(y=>+x+1*y>7?+x+1*y-7:' ').join``).join`n`

Java (JDK), 132 bytes

interface A{static void main(String[]v){for(long m=0,a=0;++a<18;System.out.printf("%"+(a>9?26-a:8+a)+"d%n",m*m))m=a>9?m/10:m*10+1;}}

Try it online!

This is a different way to do it than my first answer for Java. It is using integer arithmetics. It is even shorter, maybe the shortest possible.

Formatted version of this code:

interface A {
  static void main(String[] v) {
    for (long m = 0, a = 0; ++a < 18; System.out.printf("%" + (a > 9 ? 26 - a : 8 + a) + "d%n", m * m))
      m = a > 9 ? m / 10 : m * 10 + 1;
  }
}

Or more readable:

class A {
  public static void main(String[] v) {
    for (long m = 0, a = 0; ++a < 18; ) {
      long width = a > 9 ? 26 - a : 8 + a;
      m = a > 9 ? m / 10 : m * 10 + 1;
      System.out.printf("%" + width + "d%n", m * m);
    }
  }
}

StupidStackLanguage, 468 bytes

avdqvdmffffffffqvvviifavvfblffffffflfifdflavvfbfffffflfififdfdflavvfbffffflfifififdfdfdflavvfbfffflfififififdfdfdfdflavvfbffflfifififififdfdfdfdfdflavvfbfflfififififififdfdfdfdfdfdflavvfbflfifififififififdfdfdfdfdfdfdfavvfbfififififififififdfdfdfdfdfdfdfdflavvfbflfifififififififdfdfdfdfdfdfdflavvfbfflfififififififdfdfdfdfdfdflavvfbffflfifififififdfdfdfdfdflavvfbfffflfififififdfdfdfdflavvfbffffflfifififdfdfdflavvfbfffffflfififdfdflavvfbffffffflfifdflavvfbfffffffflf

Try it online!

Explanation

Super simple, we just create a space and the number 1, then every line just increment and print then decrement and print

Perl 5, 36 bytes

s/-//,say$"x$_.(1x(9-$_))**2for-8..8

Try it online!

Using 11 * 11 = 121, 111 * 111 = 12321, 1111 * 1111 = 1234321, ...

REXX, 86 chars

x='         12345678987654321 ' 
do n=1 to 17                    
m=9-abs(9-n)                    
say substr(x,m,10)right(x,m)    
end  

Complete program. No real tricks used except some smooshing and keeping to a single loop. I'm assuming EOL doesn't count.

Java, 156 chars

interface A{static void main(String[]v){for(int x,a=-9;++a<9;System.out.printf("%9s%s%n","123456789".substring(0,9-x),"87654321".substring(x)))x=a>0?a:-a;}}

Formatted version of this solution:

interface A {
  static void main(String[] v) {
    for (int x, a = -9; ++a < 9; System.out.printf("%9s%s%n", "123456789".substring(0, 9 - x), "87654321".substring(x)))
      x = a > 0 ? a : -a;
  }
}

Java code is generally longer. This is my shortest solution so far.

It does not produce additional whitespace in the right side as another solution does, but it is a little bit longer.

///, 139 bytes

/~/////;/3&~|/*#~*/5432~&/2^~^/#!~%/$56~$/1234~#/1
~@/  ~!/@@/!!^@ 1&@12; $;$|@ %|@%76| %7876|%789876| %7876|@%76|@ %54;$54; $;@12;@ 1&!#

Try it online!

Perl 5, 50 bytes

s/-//,$==8-$_,say$"x$_,map$=+1-abs,-$=..$=for-8..8

Try it online!

Java, 155 chars

interface A{static void main(String[]a){var o=System.out;for(int i=-9,j,k;++i<9;o.println())for(j=-9;++j<9;)o.print((k=(i<0?-i:i)+(j<0?-j:j))>8?" ":9-k);}}

-7 chars thanks to a kind commenter.

C# (Visual C# Interactive Compiler), 103 bytes

for(int i=1,j=1;i>0;i+=j=i>8?-1:j)Write($"{"123456789".Substring(0,i),9}"+"87654321n".Substring(9-i));

Try it online!

Definitely not the shortest answer, and this interpreter was published after the question was posted. But I did not see a C# answer. When using the regular C# compiler, the print statement is much longer: System.Console.Write. Also, string interpolation was not a thing when this question was posted.

Tcl, 99 bytes

time {puts [string repe { } [expr abs([incr i]-9)]][expr [string repe 1 [expr 9-abs($i-9)]]**2]} 17

Try it online!

JavaScript, 84 bytes

Saved many bytes thanks to ETHProductions

[...s="12345678987654321"].map(x=>"".padEnd(9-x)+s.slice(0,x-1)+s.slice(-x)).join`
`

Try it online!

Powershell, 45 bytes

1..9+8..1|%{' '*(9-$_)+-join(1..$_+$_..1|gu)}

2 solutions with 44 bytes were proposed in the comments to the Iszi's post

C - 118 characters

x,c,i=1,j;main(){for(;i<20;){x=10;for(j=1;j<20;){j++<10?x--:x++;printf(x>c?" ":"%d",x);}i++<10?c++:c--;printf("n");}}

My very first code golf! I still need to get into the golfing mindset, I know there is probably a lot I can do differently.

Edit: Try it online!

R, 62 bytes

for(i in c(1:9,8:1))cat(rep(" ",9-i),1:i,(i:1)[-1],"
",sep="")

Try it online!

MATL, 18 bytes

9Zv-9Zvq!-t0>48*+c

Try it on MATL Online

9Zv - symmetric range from [1:9 8:-1:1]

-9Zv - reverse symmteric range, [9:-1:1 2:9]

q! - decrement that to [8:-1:0 1:8] and transpose to vertical

- - broadcast subtract - matrix of each value in the second range subtracted from each value in the first range

t0> - duplicate that and get a logical matrix of 1s where it's > 0, 0s elsewhere

48* - multiply by 48 to change 1s to 48s (ASCII '0')

+ - add that to the original matrix

c - convert to char and implicitly display

brainfuck, 158 154 bytes

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

Try it online!

[spaceMem, spaceCount, space, lf, "0", numCount, numMem]
++++++++[->+>++++>+>++++++<<<<]>>>++<<+
[ for spaceCount
    -[- for spaceCount minus 1
        <+ inc spaceMem
        >>. print space
        < go to spaceCount
    ] 
    <[->+<]> fetch spaceCount from spaceMem
    >>>+ inc number
    >[- for numCount
        >+  inc numMem
        <<.+ print and inc number
        >   go to numCount
    ]
    >+[- for numMem plus 1
        <+ inc numCount
        <.- print and decrement number
        >>  go to numMem
    ]
    <<<. print lf
    <<      go to spaceCount
]
>>>>-[      for numCount
    <<<<+[- for spaceCount plus 1
        <+ inc spaceMem
        >>. print space
        < go to spaceCount
    ] 
        <[->+<]> fetch spaceCount from spaceMem
    >>> inc number
    >[- for numCount
        >+  inc numMem
        <<+. print and inc number
        >   go to numCount
    ]
    >-[- for numMem minus 1
        <+ inc numCount
        <-. print and decrement number
        >>  go to numMem
    ]
    <<-     decrement number
    <. print lf
    >> go to numCount
]

Japt -R, 11 bytes

9õ_õ ¬êÃê û

Try it online!

Unpacked & How it works

9õZ{Zõ q ê} ê û

9õZ{  [1..9].map(Z=>...)
Zõ      [1..Z]
q       Join with nothing; ["1", "12", ..., "123...9"]
ê       Make palindrome;   ["1", "121", ..., "123...9...321"]
}
ê     Make palindrome on the array
û     Center-pad each element to the longest one
      `-R` joins with newline
      implicit output

I like the õ_õ emoji.

Canvas, 7 bytes

９｛Ｒ⇵］↶┼

Try it here!

Explanation:

9{R⇵]↶┼  
9{  ]    map over i = 1..9
  R        push a range 1..i
   ⇵       and reverse that array
         in Canvas, an array of arrays is a list of lines,
          and the inner arrays are joined together
     ↶   rotate anti-clockwise
      ┼  and quad-palindromize with 1 overlap

I've just fixed a couple built-ins to make a 6 byte version possible:

９｛Ｒ］／┼

Try it here!

9{R]/┼
9{ ]    map over 1..9
  R       create a range 1..i
    /   pad with a diagonal of spaces
     ┼  and quad-palindromize

Kotlin, 125 bytes

fun d(){for(i in 0..307){val v=Math.abs(i%18-9)+Math.abs(i/18-8)
print(if(i%18!=0)if(v>8)' '
else(57-v).toChar()
else 'n')}}

Try it online!

Stax, 10 bytes

▌┼î▲░ò╝╪.¢

Run and debug it

Attache, 49 bytes

Print@Join=>Table[{If[_%9=_,9-_,sp]}@`+,Abs!-8:8]

Try it online!

Simply tables (that is, applies a function like a multiplication table) the function which adds two numbers (from [8, 7, 6, 5, ..., 5, 6, 7, 8]), checks if they are less than 9, then yields 9 - that number if so, otherwise a space. Then, prints each row accordingly.

///, 124 bytes

/(/# '87#//'/"67//&/123//%/43!//$/  //#/654321
//"/&45//!/21
$ /$$$$1
$$$ 1!$ &!$&% "%"#$'('8987($'65%"65% "%$&%$ &!$$1!$$ 1

Try it online!

Shorter than this answer by 109 bytes! Just applies a bunch of substitutions with whitespace and common numbers.

Canvas, 14 bytes

９｛：９∔ ×；Ｒ］∑９ｎ┼

Try it here!

Python 2, 72 bytes

z="123456789"
for i in range(9)+range(8)[::-1]:print"% 8s"%z[:i]+z[i::-1]

Try it online!

T-SQL, 116 115 bytes

DECLARE @ INT=8,@d INT=-1a:PRINT SPACE(@)+STUFF('12345678987654321',9-@,2*@,'')IF @=0SET @d=1SET @+=@d If @<9GOTO a

Pure procedural counter and loop, not very "SQL"-like, but the best I could come up with using what SQL offers. Formatted:

DECLARE @ INT=8, @d INT=-1
a:
    PRINT SPACE(@)+STUFF('12345678987654321',9-@,2*@,'')
    IF @=0 SET @d=1
    SET @+=@d
If @<9 GOTO a

Cuts a length out of a hard-coded string using STUFF. Would work just as easily with any other set of characters.

05AB1E, 9 bytes

žh¦η€ûû.c

Try it online!

///, 132 bytes

/T/123//F/T45//f/4321//S/  //s/S //N/
s//a/F678//b/765f//A/1Ns 121Ns/ssSAT21NSTfN FfNF65f
SF6b
 ab
a98b
 ab
SF6bNF65fN FfNSTfNsT2AS1

Try it online!

Yabasic, 89 bytes

Anonymous function that takes no input and outputs to STDOUT.

For i=-8To 8
For j=-8To 9
k=Abs(i)+Abs(j)
If k>8Then?" ";
Else?Chr$(57-k);
Fi
Next
?
Next

Try it online!

APL (Dyalog Classic), 21 bytes

{⊃⍵/⍕⍵}¨9-+/↑|8-⍳2⍴17

Try it online!

Jelly,  22 18 12  11 bytes

9ŒḄr1z⁶ṚŒḄY

Try it online!

Done alongside caird coinheringaahing in chat.

How it works

9ŒḄr1z⁶ṚŒḄY - Full program.

9ŒḄ         - The list [1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1].
   r1       - Generate [N, N-1, ..., 1] for each N in ^.
     z⁶     - Zip; transpose ^ with filler spaces.
       Ṛ    - Reverse.
        ŒḄ  - Palindromize without vectorization.
          Y - Join by newlines.

Saved 6 bytes thanks to Leaky Nun!

Recursiva, 31 22 bytes

{pB9'P+*" "- 9}J""WpB}

Try it online!

Explanation:

{pB9'P+*" "- 9}J""WpB}
{                       - For each
 p                      - palindromize [1,2,..9,..1]
  B9                    - range [1,2...9] 
    '                   - Iteration command begin
     P                  - Print
      +                 - concatenate
       *" "- 9 }        - appropriate number of spaces
                J""WpB} - obtain number string
                J""     - Join with nothing '123454321'
                   W    - map each element as string ['1','2'..'5','2','1']
                    p   - palindromize [1,2,..5,..2,1]
                     B} - range [1,2,3,4,5] 

Gaia, 9 bytes

9┅…ṫ¦€|ṫṣ

Explanation

9┅         Push [1 2 3 4 5 6 7 8 9]
  …        Prefixes; push [[1] [1 2] [1 2 3] ... [1 2 3 4 5 6 7 8 9]]
   ṫ¦      Palindromize each row: e.g. [1 2 3 4] -> [1 2 3 4 3 2 1]
     €|    Centre-align the rows, padding with spaces
       ṫ   Palindromize the rows
        ṣ  Join with newlines

Python 2, 59 bytes

i=8;exec'print" "*abs(i)+`int("1"*(9-abs(i)))**2`;i-=1;'*17

Try it online!

Japt -R, 19 18 17 16 11 bytes

9õõ ®¬êÃê û

Test it

PHP, 99 91 81 byte

for($y=17;$y--;print"
")for($x=17;$x--;)echo($d=9-abs($x-8)-abs($y-8))>0?$d:" ";

run in command line

php -r "CODE_ABOVE"

18 bytes saved by Jörg Hülsermann

k, 37 bytes

r:{x,1_|x};`0:`c$r@r'|8{32,-1_x}49+!9

Try it online.

Octave, 38 bytes

x=abs(-8:8);m=x+x';m(m>8)=25;[57-m,'']

To make it work in MATLAB too, you'd need to write x=ndgrid(abs(-8:8));m=x+x';m(m>8)=25;[57-m,''] or x=meshgrid(abs(-8:8));m=x+x';m(m>8)=25;[57-m,'']

Matlab 227 223 221 209 208 bytes

z=@cellfun;
C=z(@(a,b,c)[a b c fliplr([a b])],[mat2cell(repelem(' ',36),1,8:-1:1),{''}],[{''},mat2cell(nonzeros(tril(repmat(49:56,8,1))')',1,1:8)],num2cell(49:57),'un',0)';
z(@(c)disp(c),[C;flipud(C(1:end-1))])

///, 233 bytes

        1
       121
      12321
     1234321
    123454321
   12345654321
  1234567654321
 123456787654321
12345678987654321
 123456787654321
  1234567654321
   12345654321
    123454321
     1234321
      12321
       121
        1

Try it online!

Yay.

Deadfish, 1446 bytes

iisiisddddooooooooiiiiiiiiiiiiiiiiiodddddddddddddddddddddddddddddddddddddddoddddsddddoooooooiiiiiiiiiiiiiiiiioiododddddddddddddddddddddddddddddddddddddddoddddsddddooooooiiiiiiiiiiiiiiiiioioiodododddddddddddddddddddddddddddddddddddddddoddddsddddoooooiiiiiiiiiiiiiiiiioioioiododododddddddddddddddddddddddddddddddddddddddoddddsddddooooiiiiiiiiiiiiiiiiioioioioiodododododddddddddddddddddddddddddddddddddddddddoddddsddddoooiiiiiiiiiiiiiiiiioioioioioiododododododddddddddddddddddddddddddddddddddddddddoddddsddddooiiiiiiiiiiiiiiiiioioioioioioiodododododododddddddddddddddddddddddddddddddddddddddoddddsddddoiiiiiiiiiiiiiiiiioioioioioioioiododododododododddddddddddddddddddddddddddddddddddddddodddsoioioioioioioioiodododododododododddddddddddddddddddddddddddddddddddddddoddddsddddoiiiiiiiiiiiiiiiiioioioioioioioiododododododododddddddddddddddddddddddddddddddddddddddoddddsddddooiiiiiiiiiiiiiiiiioioioioioioiodododododododddddddddddddddddddddddddddddddddddddddoddddsddddoooiiiiiiiiiiiiiiiiioioioioioiododododododddddddddddddddddddddddddddddddddddddddoddddsddddooooiiiiiiiiiiiiiiiiioioioioiodododododddddddddddddddddddddddddddddddddddddddoddddsddddoooooiiiiiiiiiiiiiiiiioioioiododododddddddddddddddddddddddddddddddddddddddoddddsddddooooooiiiiiiiiiiiiiiiiioioiodododddddddddddddddddddddddddddddddddddddddoddddsddddoooooooiiiiiiiiiiiiiiiiioiododddddddddddddddddddddddddddddddddddddddoddddsddddooooooooiiiiiiiiiiiiiiiiiodddddddddddddddddddddddddddddddddddddddo

A more human friendly spaced version:

iisiisdddd oooooooo iiiiiiiiiiiiiiiii o dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooooooo iiiiiiiiiiiiiiiii oiodo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooooo iiiiiiiiiiiiiiiii oioiododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooooo iiiiiiiiiiiiiiiii oioioiodododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooo iiiiiiiiiiiiiiiii oioioioiododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooo iiiiiiiiiiiiiiiii oioioioioiodododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oo iiiiiiiiiiiiiiiii oioioioioioiododododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd o iiiiiiiiiiiiiiiii oioioioioioioiodododododododo dddddddddddddddddddddddddddddddddddddddo
ddds oioioioioioioioiododododododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd o iiiiiiiiiiiiiiiii oioioioioioioiodododododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oo iiiiiiiiiiiiiiiii oioioioioioiododododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooo iiiiiiiiiiiiiiiii oioioioioiodododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooo iiiiiiiiiiiiiiiii oioioioiododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooooo iiiiiiiiiiiiiiiii oioioiodododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooooo iiiiiiiiiiiiiiiii oioiododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooooooo iiiiiiiiiiiiiiiii oiodo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooooooo iiiiiiiiiiiiiiiii o dddddddddddddddddddddddddddddddddddddddo

Groovy, 62, 57 chars

((1..9)+(8..1)).any{println' '*(9-it)+('1'*it as int)**2}

old version:

((1..9)+(8..1)).any{println"${('1'*it as int)**2}".center(17)}

explanation: we create a list [1,2,...,9,8,7,..,1]. Within the closure we create strings '1', '11', '111,..., convert them to numbers, run power of two and center.

Common Lisp SBCL, 145 bytes

(do((b 1)(l'(1 2 3 4 5 6 7 8 9))(i 1(+ i b)))((= i 0))(format t"~17:@<~v{~a~}~v{~a~}~>~%"i l i(reverse(subseq l 0(1- i))))(if(> i 8)(setf b -1)))

It is worse than the other Common Lisp solution, but it works and I think it is different enough to be posted.

Explanation

i is incremented up to 9 and then decremented to 0 (when i=9 sign of b is changed effectively changing addition to subtraction - this avoids second loop.

In each line I print numbers: `123...i(i-1)...1 using loops of format function (for first loop I use list '(1 2 3 4 5 6 7 8 9) and for decrementing loop I use reversed subsequence of this list. Text is then centered.

I didn't notice that numbers are squares of 11..11. As far as this solution is concerned the diamond could be made out of letters or (! @ # ...) (for that you would need to change ~a to ~c in format function.

Clojure, 191 179 bytes

#(loop[[r & s](range 18)h 1](print(apply str(repeat(if(< r 8)(- 8 r)(- r 8)) )))(doseq[m(concat(range 1 h)(range h 0 -1))](print m))(println)(if s(recur s((if(< r 8)inc dec)h))))

-12 bytes by changing the outer doseq to a loop, which allowed me to get rid of the atom (yay).

A double "for-loop". The outer loop (loop) goes over each row, while the inner loop (doseq) goes over each number in the row, which is in the range (concat (range 1 n) (range n 0 -1)), where n is the highest number in the row.

(defn diamond []
  (let [spaces #(apply str (repeat % " "))] ; Shortcut function that produces % many spaces
    (loop [[row-n & r-rows] (range 18) ; Deconstruct the row number from the range
           high-n 1] ; Keep track of the highest number that should appear in the row
      (let [top? (< row-n 8) ; Are we on the top of the diamond?
            f (if top? inc dec) ; Decided if we should increment or decrement
            n-spaces (if top? (- 8 row-n) (- row-n 8))] ; Calculate how many prefix-spaces to print
        (print (spaces n-spaces)) ; Print prefix-spaces
        (doseq [m (concat (range 1 high-n) (range high-n 0 -1))] ; Loop over the row of numbers
          (print m)) ; Print the number
        (println)

        (if r-rows
          (recur r-rows (f high-n)))))))

Due to a bug in the logic in my first attempt (accidentally inserting the prefix-spaces between each number instead), I managed to get this:

1
1       2       1
1      2      3      2      1
1     2     3     4     3     2     1
1    2    3    4    5    4    3    2    1
1   2   3   4   5   6   5   4   3   2   1
1  2  3  4  5  6  7  6  5  4  3  2  1
1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
12345678987654321
1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1
1  2  3  4  5  6  7  8  9  8  7  6  5  4  3  2  1
1   2   3   4   5   6   7   8   7   6   5   4   3   2   1
1    2    3    4    5    6    7    6    5    4    3    2    1
1     2     3     4     5     6     5     4     3     2     1
1      2      3      4      5      4      3      2      1
1       2       3       4       3       2       1
1        2        3        2        1
1         2         1

Not even correct ignoring the obvious bug, but it looked cool.

tcl, 122 110

proc P i {puts [format %[expr 8+$i]s [expr [string repe 1 $i]**2]]}
time {P [incr i]} 9
time {P [incr i -1]} 8

demo

SmileBASIC, 65 bytes

FOR I=-8TO 8A=ABS(I)Q=VAL("1"*(9-A))?" "*A;Q*Q
NEXT
LOCATE 16,8?1

I used the fact that 11*11=121, 111*111=12321, etc. Unfortunately, 12345678987654321 can't be stored as a 64 bit float, so I had to add the last 1 separately, adding 14 bytes.

Excel VBA, ₈₁ 76 Bytes

Anonymous VBE immediate window function that takes no input and outputs to the VBE Immediate Window

For i=-8To 8:j=Abs(i):r=Mid(987654321,j+1):?Spc(j)StrReverse(r)Mid(r,2):Next

Old Version, 81 Bytes

For i=-9To 8:For j=-8To 9:k=Abs(i)+Abs(j):l=l &IIf(k>8," ",9-k):Next:?l:l="":Next

Vim, 62 39 38 34 keystrokes

Thanks to @DJMcMayhem for saving a ton of bytes

Thanks to @AaronMiller for saving 4 bytes by generating 12345678987654321 in a different way

9i1<ESC>|C<C-r>=<C-r>"*<C-r>"
<ESC>qqYP9|xxI <ESC>YGpHq7@q

Try it online!

Explanation:

9i1<ESC>                            Write 9 '1's, leaving the cursor at the end of the line
        |C                          Go to the first column and cut all that's to the right of the cursor (implictly into register "), entering insert mode
          <C-r>=                    Enter an expression:
                <C-r>"*<C-r>"        The contents of register " multiplied with itself
                             <CR>   Evaluate it, yielding 12345678987654321

qq                                  Start recording into register q
  YP                                Yank this entire line and Paste above
    9|                              Go to the 9th column
      xx                            Delete character under cursor twice
        I <ESC>                     Go to the beginning of the line and insert a space and enter normal mode
               Y                    Yank this entire line
                G                   Go to the last line
                 p                  Paste in the line below
                  H                 Go to the first line
                   q                End recording
                    7@q             Repeat this 7 times

APL (Dyalog Classic), 20 19 bytes

(⍉⊢⍪1↓⊖)⍣2⌽↑,⍨1↓⎕d

Try it online!

⎕d are the digits '0123456789'

1↓ drop the first ('0')

,⍨ swapped catenate scan, i.e. the reversed prefixes '1' '21' '321' ... '987654321'

↑ mix into a matrix padded with spaces:

1
21
321
...
987654321

⌽ reverse the matrix horizontally

(...)⍣2 do this twice:

⍉⊢⍪1↓⊖ the transposition (⍉) of the matrix itself (⊢) concatenated vertically (⍪) with the vertically inverted matrix (⊖) without its first row (1↓)

Ruby, 55

puts (-8..8).map{|i|[?s*a=i.abs,(?1*(9-a)).to_i**2]*''}

Output:

irb(main):342:0> puts (-8..8).map{|i|[?s*a=i.abs,(?1*(9-a)).to_i**2]*''}
        1
       121
      12321
     1234321
    123454321
   12345654321
  1234567654321
 123456787654321
12345678987654321
 123456787654321
  1234567654321
   12345654321
    123454321
     1234321
      12321
       121
        1

JavaScript, 170 bytes

My first code golf :)

Golfed

a="";function b(c){a+=" ".repeat(10-c);for(i=1;i<c;i++)a+=i;for(i=2;i<c;i++)a+=c-i;a+="n";}for(i=2;i<11;i++)b(i);for(i=9;i>1;i--)b(i);document.write("<pre>"+a+"</pre>");

Ungolfed

var str = "";
function row(line) {
    str += " ".repeat(10 - line);
    for (var i = 1; i < line; i++) {
        str += i;
    }
    for (var i = 2; i < line; i++) {
        str += line - i;
    }
    str += "n";
}
for (var line = 2; line < 11; line++) {
    row(line);
}
for (var line = 9; line > 1; line--) {
    row(line);
}
document.write("<pre>" + str + "</pre>");

Ruby 59

(-8..8).map{|i|puts' '*i.abs+"#{eval [?1*(9-i.abs)]*2*?*}"}

Befunge-93, 155 chars

9:v:<,+55<v5*88<v-9:$_68v
> v>     ^>3p2vpv  -1<!  *
, 1^  2p45*3+9<4:    ,:  +
g -^_75g94+4pg7^!    +^ ,<
1 : ^ `0    :-1$_:68*^$
^1_$:55+-0>:#$1-#$:_^

Try it online!

It could definitely be golfed more, but it's my first Funge program and my head already hurts. Had a lot of fun, though

Haskell, 77 bytes

r=[-8..8]
f n|n<1=" "|1>0=show n
mapM putStrLn[[9-abs x-abs y|x<-r]>>=f|y<-r]

05AB1E, 17 15 10 bytes

9LJ.pû€û.c

Try it online!

Charcoal, 13 bytes

Ｆ⁹«ＧX⁻⁹ιＩ⁺ι¹↓

Try it online!

How?

Draws nine, successively smaller, concentric number-diamonds on top of each other:

Ｆ⁹«   Loop ι from 0 to 8:
ＧX     Draw a (filled) polygon with four equilateral diagonal sides
⁻⁹ι      of length 9-ι
Ｉ⁺ι¹    using str(ι+1) as the character
↓       Move down one space before drawing the next one

Bash, 126 109 87 chars

87:

q()(printf %$[9+$1%9]s\n $[$2*$2];[ 7 -lt $1 ]||(q $[$1+1] ${2}1;q $[$1+9] $2))
q 0 1

As it usually goes, changing from iterative to recursive solution helps us win additional bytes.

Meaning of parameters to q:

$1 How much to remove from 8 to get the number of spaces in the beginning. Note value modulo 9 counts here (actual value is also a hint to quit recursion).

$2 The current chain of 1s to be squared and output by printf.

The modus operandi is:

1. output the sequence (ie. if $2 is 11111, output 123454321)

2. (if not yet at 12..9..21 - the recursive step)

   2.1. output the next sequence (here: 111111 > $2 , output 12345654321

   2.2. output the sequence once again (123454321).

In the step 2.2 , we pass (indent value + 9) instead of indent value however, so that the algoritm "knows" we are printing the row for the second time. Without this, the [ 7 -lt $1 ] would be false, causing us to retrigger the recursive step 1. This would never finish then.

The recursion goes like this:

q 0 1:                          1
 q 1 11:                       121
  q 2 111:                    12321
   q 3 1111:                 1234321
    q  4 11111:             123454321
     q  5 111111:          12345654321
      q  6 1111111:       1234567654321
       q  7 11111111:    123456787654321
        q  8 111111111: 12345678987654321
        q 16 11111111:   123456787654321
       q 15 1111111:      1234567654321
      q 14 111111:         12345654321
     q 13 11111:            123454321
    q 12 1111:               1234321
   q 11 111:                  12321
  q 10 11:                     121
 q  9 1:                        1

p()(printf "%$[8+i]sn" $[k*k])
k=;for i in `seq 9`;do k+=1;p;done;for i in `seq 8 -1 1`;do k=${k:1};p;done;

"k+=1" is much cheaper as k=$[10*k+1] , and for k being a string of ones it's the same. Same goes for ${k:1} and $[k/10] .

126:

p() (printf "%$[$1+i]sn" $[k*k];)
k=1;for i in `seq 8`;do p 8;k=$[10*k+1];done;for i in `seq 8 -1 0`;do p 9;k=$[k/10];done;

I guess there may be even shorter solution, but weather is glorious, I can't stand sitting in front of computer any more :).

CJam, 31 27 bytes

CJam is a lot newer than this challenge, so this answer is not eligible for being accepted. This was a neat little Saturday evening challenge, though. ;)

8S*9,:)+9*9/2%{_W%1>+z}2*N*

Test it here.

The idea is to form the upper left quadrant first. Here is how that works:

First, form the string " 123456789", using 8S*9,:)+. This string is 17 characters long. Now we repeat the string 9 times, and then split it into substrings of length 9 with 9/. The mismatch between 9 and 17 will offset every other row one character to the left. Printing each substring on its own line we get:

        1
23456789 
       12
3456789  
      123
456789   
     1234
56789    
    12345
6789     
   123456
789      
  1234567
89       
 12345678
9        
123456789

So if we just drop every other row (which conveniently works by doing 2%), we obtain one quadrant as desired:

        1
       12
      123
     1234
    12345
   123456
  1234567
 12345678
123456789

Finally, we mirror this twice, transposing the grid in between to ensure that the two mirroring operations go along different axes. The mirroring itself is just

_      "Duplicate all rows.";
 W%    "Reverse their order.";
   1>  "Discard the first row (the centre row).";
     + "Add the other rows.";

Lastly, we just join all lines with newlines, with N*.

Excel, 24 chars (cheating?)

1. In cell B2 put =MAX(A2,B1,B3,C2)-1 (19 chars)
2. Fill down to B18 and right to R18.
3. Select B2 through B18 and Format:Cells:Number:Custom: 0; (2 chars)
4. Enter a 8.9 in J10 (3 char)
5. Ignore all complaints about circular references throughout the process. Use Preferences:Calculation to allow iteration if it is not already allowed.
6. Select columns B through R and resize their width to make square boxes.
7. Profit?

Note: Some might call step 2 cheating as it creates 19 to 23 characters in every cell for a total of over 6000 chars. If you really want to count it that way, you would be better off not putting the formula into those squares that are to remain blank. In that case, you can use a 9 in J10 and you don't need the custom formatting. The total character count would then be just over 3000.

APL, 24 chars

⍉⊃{⍕⍵↑⍨⍵>0}¨9-∘.+⍨|9-⍳17

Tested in Nars2000 and Dyalog (requires ⎕ML←3 in the latter.)

Explanation

                     ⍳17    starting with the naturals up to 17
                  |9-       generate the numbers from 8 to 0 and back to 8
              ∘.+⍨          make a table of their sum (with 0 in the middle)
            9-              turn it into a diamond with 9 in the middle
  {       }¨                for each number
    ⍵↑⍨⍵>0                  keep it only if it's positive
   ⍕                        then convert the result, if any, to a string
⍉⊃                          disclose the nested array and adjust the dimensions

The last step transposes the result, whose shape is 17 17 1 (because of the disclose ⊃ of nested strings) into 1 17 17, which gets printed like a plain 17 17.

Output

⍉⊃{⍕⍵↑⍨⍵>0}¨9-∘.+⍨|9-⍳17
        1        
       121       
      12321      
     1234321     
    123454321    
   12345654321   
  1234567654321  
 123456787654321 
12345678987654321
 123456787654321 
  1234567654321  
   12345654321   
    123454321    
     1234321     
      12321      
       121       
        1        

C++ 223 Byte

#include <iostream>
using std::cout;using std::size_t;int main(){for(int a=0;a<2;++a)for(size_t b=1+a*7;b<10-a;((a!=1)?++b:--b)){size_t c=9-b;for(;c-->0;)cout<<" ";for(c=1;c<b;)cout<<c++;for(c=b;0<c;)cout<<c--;cout<<'n';}}

Ungolfed:

#include <iostream>
using std::cout; //for not having to type std::cout over and over again
using std::size_t; //for not having to type std::size_t over and over again

int main()
{
    for(int a = 0; a < 2; ++a)
        for(size_t b=1+a*7; b<10-a; ((a!=1)?++b:--b))
        {     //either count up to nine or down from nine
            size_t c = 9-b; //space count we need
            for(; c-- > 0;)
                cout << " ";
            for(c = 1; c < b;) //set c to the counter that will be print
                cout << c++; //post-crement :)
            for(c = b; 0 < c;) //count backwards
                cout << c--; //post-decrement :)
            cout << 'n'; //line is done
        }
}

J, 29 26 24 23 22 21 chars

,.(0&<#":)"+9-+/~|i:8

Thanks to FUZxxl for the "+ trick (I don't think I've ever used u"v before, heh).

Explanation

                  i:8  "steps" vector: _8 _7 _6 ... _1 0 1 ... 7 8
                 |     magnitude
              +/~      outer product using +
            9-         inverts the diamond so that 9 is in the center
  (      )"+           for each digit:
      #                  copy
   0&<                   if positive then 1 else 0
       ":                copies of the string representation of the digit
                         (in other words: filter out the strictly positive
                          digits, implicitly padding with spaces)
,.                     ravel each item of the result of the above
                       (necessary because the result after `#` turns each
                        scalar digit into a vector string)

PowerShell (2 options): 92 84 45 bytes

1..8+9..1|%{' '*(9-$_)+[int64]($x='1'*$_)*$x}
1..9+8..1|%{' '*(9-$_)+[int64]($x='1'*$_)*$x}

Thanks to Strigoides for the hint to use 1^2,11^2,111^2...

Shaved some characters by:

• Eliminating $w.
• Nested the definition of $x in place of its first use.
• Took some clues from Rynant's solution:
  • Combined the integer arrays with + instead of , which allows elimination of the parenthesis around the arrays and a layer of nesting in the loops.
  • Used 9-$_ to calculate the length of spaces needed, instead of more complicated maths and object methods. This also eliminated the need for $y.

Explanation:

1..8+9..1 or 1..9+8..1 generates an array of integers ascending from 1 to 9 then descending back to 1.

|%{...} pipes the integer array into a ForEach-Object loop via the built-in alias %.

' '*(9-$_)+ subtracts the current integer from 9, then creates a string of that many spaces at the start of the output for this line.

[int64]($x='1'*$_)*$x defines $x as a string of 1s as long as the current integer is large. Then it's converted to int64 (required to properly output 111111111² without using E notation) and squared.

PowerShell, 49 48

1..9+8..1|%{"  "*(9-$_)+(1..$_+($_-1)..0|?{$_})}

APL (40)

r←{⍵,1↓⌽⍵}
{⎕←⍵,⍨' '⍴⍨(2×10-⌈/⍵)}¨r¨r⍳¨⍳9

I guess I'm not beating marinus. :p

GolfScript (27 chars)

17,{8-abs' '*1`9*1$,>~.*n}/

or

17,{8-abs' '*.1`9*+9<~.*n}/

Both work by building a suitable repunit as a string and then converting to int and squaring to get a Demlo number.

JavaScript, 81

for(i=9;--i+9;console.log(s))for(j=9;j;s=j--^9?k>0?k+s+k:" "+s:k+"")k=i<0?j+i:j-i

Javascript 88

for(i=9,a=Math.abs;--i>-9;console.log(o))for(j=9,o='';j-->-9;)o+=(n=9-a(i)-a(j))>0?n:' '

Mathematica 55 50 45 41 38

(10^{9-Abs@Range[-8,8]}-1)^2/81//Grid

Grid[(10^Array[{9}-Abs[#-9]&,17]-1)^2/81]

Javascript, 137

With recursion:

function p(l,n,s){for(i=l;i;s+=" ",i--);for(i=1;i<=n;s+=i++);for(i-=2;i>0;s+=i--);return(s+="n")+(l?p(l-1,n+1,"")+s:"")}alert(p(8,1,""))

First time on CG :)

Or 118

If I can find a JS implementation that executes 111111111**2 with higher precision.
(Here: 12345678987654320).

a="1",o="n";for(i=0;i<9;i++,o+="         ".substr(i)+a*a+"n",a+="1");for(i=8;i;i--)o+=o.split("n")[i]+"n";alert(o)

Scala - 86 characters

val a="543210/.-./012345";for(i<-a){for(j<-a;k=99-i-j)print(if(k<1)" "else k);println}

Python 2, 60 59

for n in`111111111**2`:print`int('1'*int(n))**2`.center(17)

Abuses backticks and repunits.

Raku, 42 41 38 35 chars

say " "x 9-$_,(1 x$_)²for 1…9…1

Try it online!

Javascript, 114

My first entry on Codegolf!

for(l=n=1;l<18;n-=2*(++l>9)-1,console.log(s+z)){for(x=n,s="";x<9;x++)z=s+=" ";for(x=v=1;x<2*n;v-=2*(++x>n)-1)s+=v}

If this can be shortened any further, please comment :)

C, 79 chars

v;main(i){for(;i<307;putchar(i++%18?v>8?32:57-v:10))v=abs(i%18-9)+abs(i/18-8);}

Python, 65

for i in map(int,str(int('1'*9)**2)):print' '*(9-i),int('1'*i)**2

Groovy 77 75

i=(-8..9);i.each{a->i.each{c=a.abs()+it.abs();print c>8?' ':9-c};println""}

old version:

(-8..9).each{a->(-8..9).each{c=a.abs()+it.abs();print c>8?' ':9-c};println""}

Javascript, 129* 126

for(i=1;i<18;i++){s="";a=Math.abs(9-i);for(j=0;j<a;j++)s+=" ";for(k=a+1;k<=9;k++)s+=k-a;for(l=8;l>a;l--)s+=l-a;console.log(s)}

Includes suggestion from Shmiddty in comments. Original preserved below:

for(i=1;i<18;i++){s="";a=Math.abs(9-i);for(j=0;j<a;j++){s+=" "}for(k=a+1;k<=9;k++){s+=k-a}for(l=8;l>a;l--){s+=l-a}console.log(s)}

I'm sure this could be condensed further, but darned if I know how. :P

Perl, 43+1

adding +1 for -E which is required for say

say$"x(9-$_).(1x$_)**2for 1..9,reverse 1..8

edit: shortened a bit

VBA, 197

for i=1 to 9:?string(27-i*3,32)l;:for j=1 to i:?j;:next:for j=i-1 to 1 step -1:?j;:next:?:next:for i=8 to 1 step -1:?string(27-i*3,32)l;:for j=1 to i:?j;:next:for j=i-1 to 1 step -1:?j;:next:?:next

if vba didn't automatically add a space for the + sign it doesn't print, perhaps I could avoid the 27-i*3 construct for making it look right

K, 59

-1'(-:'9+k,1_|k:!9)$,/'$b,1_||:'b:(-1_'a),'|:'a:1_1+!:'!10;

Java, 177 chars

public class A{public static void main(String[]v){
for(int a=-8,b,c;a<9;a++){
for(b=-8;b<9;){c=Math.abs(a)+Math.abs(b++);
System.out.print(c>8?" ":9-c);}System.out.println();}}}

correct formatting of this solution (352 chars):

public class A {
    public void main(String[] args) {
        for (int a = -8; a < 9; a++) {
            for (int b = -8; b < 9;) {
                int c = Math.abs(a) + Math.abs(b++);
                System.out.print(c > 8 ? " " : 9 - c);
            }
            System.out.println("");
        }
    }
}

Ruby, 76 69 60 54 characters

(-8..8).map{i=_1.abs;puts' '*i+"#{eval(?1*(9-i))**2}"}

(Thanks, Patrick, G B, and Slim Liser!)

Improvements welcome. :)

R, 71 characters

For the records:

s=c(1:9,8:1);for(i in s)cat(rep(" ",9-i),s[0:i],s[(i-1):0],"n",sep="")

C, 98 chars

l=9,n;p(i){for(i=18;i;putchar(i?n>l?48+n-l:32:10))n=9<--i?18-i:i;}
main(){p(l--);l&&p(main());l++;}

Common Lisp, 113 characters

(defun x(n)(if(= n 0)1(+(expt 10 n)(x(1- n)))))(dotimes(n 17)(format t"~17:@<~d~>~%"(expt(x(- 8(abs(- n 8))))2)))

First I noticed that the elements of the diamond could be expressed like so:

  1   =   1 ^ 2
 121  =  11 ^ 2
12321 = 111 ^ 2

etc.

x recursively calculates the base (1, 11, 111, etc), which is squared, and then printed centered by format. To make the numbers go up to the highest term and back down again I used (- 8 (abs (- n 8))) to avoid a second loop

APL (33 31)

A⍪1↓⊖A←A,0 1↓⌽A←⌽↑⌽¨⍴∘(1↓⎕D)¨⍳9

If spaces separating the numbers are allowed (as in the Mathematica entry), it can be shortened to 28 26:

A⍪1↓⊖A←A,0 1↓⌽A←⌽↑⌽∘⍕∘⍳¨⍳9

Explanation:

• (Long program:)
• ⍳9: a list of the numbers 1 to 9
• 1↓⎕D: ⎕D is the string '0123456789', 1↓ removes the first element
• ⍴∘(1↓⎕D)¨⍳9: for each element N of ⍳9, take the first N elements from 1↓⎕D. This gives a list: ["1", "12", "123", ... "123456789"] as strings

• ⌽¨: reverse each element of this list. ["1", "21", "321"...]

• (Short program:)

• ⍳¨⍳9: the list of 1 to N, for N [1..9]. This gives a list [[1], [1,2], [1,2,3] ... [1,2,3,4,5,6,7,8,9]] as numbers.
• ⌽∘⍕∘: the reverse of string representation of each of these lists. ["1", "2 1"...]
• (The same from now on:)
• A←⌽↑: makes a matrix from the list of lists, padding on the right with spaces, and then reverse that. This gives the upper quadrant of the diamond. It is stored in A.
• A←A,0 1↑⌽A: A, with the reverse of A minus its first column attached to the right. This gives the upper half of the rectangle. This is then stored in A again.
• A⍪1↓⊖A: ⊖A is A mirrored vertically (giving the lower half), 1↓ removes the top row of the lower half and A⍪ is the upper half on top of 1↓⊖A.

Perl 56 54 characters

Added 1 char for the -p switch.

Uses squared repunits to generate the sequence.

s//12345678987654321/;s|(.)|$/.$"x(9-$1).(1x$1)**2|eg

Python 2, 72 69 67 61

Not clever:

s=str(111111111**2)
for i in map(int,s):print'%8s'%s[:i-1]+s[-i:]

GolfScript, 33 31 30 characters

Another GolfScript solution

17,{8-abs." "*10@-,1>.-1%1>n}%

Thank you to @PeterTaylor for another char.

Previos versions:

17,{8-abs" "*9,{)+}/9<.-1%1>+}%n*

(run online)

17,{8-abs" "*9,{)+}/9<.-1%1>n}%

Mathematica 83 49 43 54 51

Print@@#&/@(Sum[k~DiamondMatrix~17,{k,0,8}]/.0->" ")

With 3 bytes saved thanks to Kelly Lowder.

Analysis

The principal part of the code, Sum[DiamondMatrix[k, 17], {k, 0, 8}], can be checked on WolframAlpha.

The following shows the underlying logic of the approach, on a smaller scale.

a = 0~DiamondMatrix~5;
b = 1~DiamondMatrix~5;
c = 2~DiamondMatrix~5;
d = a + b + c;
e = d /. 0 -> "";
Grid /@ {a, b, c, d, e}

k (64 50 chars)

-1'(::;1_|:)@:((|!9)#'" "),'$i*i:"J"$(1+!9)#'"1";

Old method:

-1',/(::;1_|:)@:((|!9)#:" "),',/'+(::;1_'|:')@:i#:,/$i:1+!9;

GolfScript, 36 chars

Assuming that this is meant as a code-golf challenge, here's a basic GolfScript solution:

9,.);-1%+:a{a{1$+7-.0>" "if}%;n}%

PHP, 92 90 characters

<?for($a=-8;$a<9;$a++){for($b=-8;$b<9;){$c=abs($a)+abs($b++);echo$c>8?" ":9-$c;}echo"n";}

Calculates and prints the Manhattan distance of the position from the centre. Prints a space if it's less than 1.

An anonymous user suggested the following improvement (84 characters):

<?for($a=-8;$a<9;$a++,print~õ)for($b=-8;$b<9;print$c>8?~ß:9-$c)$c=abs($a)+abs($b++);