AWK, 161 bytes

{if(1~$0||!$0)print"X"
for(;i<2^$1;i++){j=i
for(x=X;j;j=int(j/2))x=j%2x
x=sprintf("%0"$1"d",x)
gsub(/10/,"X-",x)
gsub(/01|0X|-1/,"-X",x)
if(x!~/0|1|XX/)print x}}

Attempt This Online!

Just for fun this will actually print all of the checkmates based on input... good up to about 16 urinals.

Vyxal, 7 bytes

⁺»‡_+TḞ

Try it Online!

Uses the same recurrence relation a(n) = a(n-2) + a(n-3)

      Ḟ # Create an infinite list
⁺»      # Starting from digits of 112
  ‡     # Generate new terms by a function
     T  # Taking in the last three terms
   _    # Pop the first
    +   # Add the other two together

05AB1E, 8 bytes

ƵBSλè₂₃+

Try it online or verify the infinite list.

Explanation:

ƵB        # Push compressed integer 112
  S       # Convert it to a list of digits: [1,1,2]
   λ      # Start a recursive environment,
    è     # to output the 0-based (implicit) input'th value
          # (which will be output implicitly as result afterwards)
ƵBS       # Starting at a(0)=1, a(1)=1, a(2)=2
          # And where every following a(n) is calculated as:
     ₂₃+  #  a(n-2) + a(n-3)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why ƵB is 112.

Haskell, 27 bytes

r=1:1:2:zipWith(+)r(tail r)

Try it online!

C (gcc), 33 bytes

f(n){return n<2?1:f(n-2)+f(n-3);}

Try it online!

Stax, 7 bytes

ÉKΦΘÄO¢

Run and debug it

Uses the recurrence relation. C(n) = C(n-2) + C(n-3)

Actually, 18 bytes

This is an Actually port of Dennis' longer Jelly answer. Golfing suggestions welcome. Try it online!

3+;╖½Lur⌠;τ╜-@█⌡MΣ

Ungolfing

         Implicit input n.
3+       Add 3. For readibility, m = n+3.
;╖       Duplicate and store one copy of m in register 0.
½Lu      floor(m/2) + 1.
r        Range from 0 to (floor(m/2)+1), inclusive.
⌠...⌡M   Map the following function over the range. Variable k.
  ;        Duplicate k.
  τ╜-      Push m-2k. Stack: [m-2k, k]
  @█       Swap k and m-2k and take binomial (k, m-2k).
            If m-2k > k, █ returns 0, which does not affect the sum() that follows.
         End function.
Σ        Sum the list that results from the map.
         Implicit return.

Perl, 35 34 bytes

Includes +1 for -p

Give input on STDIN

checkmate.pl <<< 8

checkmate.pl:

#!/usr/bin/perl -p
$\+=$b-=$.-=$\-$b*4for(++$\)x$_}{

A newly developed secret formula. Ripple update 3 state variables without the need for parallel assignments.

It is equally short (but a lot slower and taking a lot more memory) to just solve the original problem:

#!/usr/bin/perl -p
$_=grep!/XX|\B-\B/,glob"{X,-}"x$_

but that doesn't work for 0

Actually, 25 bytes

This seems a little long for a simple f(n) = f(n-2) + f(n-3) recurrence relation. Golfing suggestions welcome. Try it online!

╗211╜¬`);(+)`nak╜2╜2<I@E

Ungolfing

         Implicit input n.
╗        Save n to register 0.
211      Stack: 1, 1, 2. Call them a, b, c.
╜¬       Push n-2.
`...`n   Run the following function n-2 times.
  );       Rotate b to TOS and duplicate.
  (+       Rotate a to TOS and add to b.
  )        Rotate a+b to BOS. Stack: b, c, a+b
         End function.
ak       Invert the resulting stack and wrap it in a list. Stack: [b, c, a+b]
╜        Push n.
2        Push 2.
╜2<      Push 2 < n.
I        If 2<n, then 2, else n.
@E       Grab the (2 or n)th index of the stack list.
         Implicit return.

05AB1E, 12 bytes

XXXIGX@DŠ0@+

Explanation

XXX            # initialize stack as 1, 1, 1
   IG          # input-1 times do:
     X@        # get the item 2nd from bottom of the stack
       DŠ      # duplicate and push one copy down as 2nd item from bottom of the stack
         0@    # get the bottom item from the stack
           +   # add the top 2 items of the stack (previously bottom and 2nd from bottom)
               # implicitly print the top element of the stack after the loop

Try it online!

CJam, 20 bytes

1_2_{2$2$+}ri*;;;o];

Try it online!

Explanation

This uses the recurrence relationship shown in the OEIS page.

1_2_                   e# Push 1, 1, 2, 2 as initial values of the sequence
           ri          e# Read input
    {     }  *         e# Repeat block that many times
     2$2$              e# Copy the second and third elements from the top
         +             e# Add them
              ;;;      e# Discard the last three elements
                 o     e# Output
                  ];   e# Discard the rest to avoid implicit display

FRACTRAN, 104 93 bytes

Input is 11**n*29 and output is 29**checkmate(n).

This is mostly for fun, especially since I'm currently being outgolfed by Python, JS, and Java. Same number of bytes as PHP though :D Golfing suggestions welcome.

403/85 5/31 3/5 9061/87 3/41 37/3 667/74 37/23 7/37 38/91 7/19 5/77 1/7 1/17 1/2 340/121 1/11

Ungolfing

               At the start we have 11**n * 29
1/11           If n < 2, we remove the 11s and print 29**1
340/121        If n >= 2, we subtract two 11s (n-2) and add one 17, two 2s and one 5.
                 We now have 17**1 * 29**1 * 2**2 * 5.
                 These are the register for a, b, c at registers 17, 29, and 2.
                 5 is an indicator to start the first loop.
                 This loop will move a to register 13.
403/85 5/31    Remove the 17s one at a time, adds them to the 13 register.
                 5 and 31 reset the loop.
3/5            Next loop: moves b to a and adds b to a in register 13.
9061/87 3/41   Remove the 29s one at a time, adds them to the 17 and 13 registers.
                 3 and 41 reset the loop.
37/3           Next loop: moves c to b in register 29.
667/74 37/23   Remove the 2s one at a time, adds them to the 29 register.
                 37 and 23 reset the loop.
7/37           Next loop: moves a+b to c in register 2.
38/91 7/19     Remove the 13s one at a time, adds them to the 2 register.
                 7 and 19 reset the loop.
5/77           Move to the first loop if and only if we have an 11 remaining.
1/7 1/17 1/2   Remove the 7 loop indicator, and all 17s and 2s.
               Return 29**checkmate(n).

JavaScript (ES6) / Recursive, 30 27 bytes

Edit: saved 3 bytes thanks to Shaun H

let

f=n=>n<3?n||1:f(n-2)+f(n-3)

for(var n = 1; n < 16; n++) {
  console.log(n, f(n));
}

JavaScript (ES6) / Non-recursive 90 77 bytes

Edit: saved 13 bytes thanks to Conor O'Brien and Titus

let f =

n=>[...Array(k=1<<n)].map((_,i)=>r+=!(i&(x=i>>1|i+i))&&((i|x)&~k)==k-1,r=0)|r

for(var n = 1; n < 16; n++) {
  console.log(n, f(n));
}

><>, 25+2 = 27 bytes

211rv
v!?:<r@+@:$r-1
>rn;

Needs the input to be present on the stack at program start, so +2 bytes for the -v flag. Try it online!

The first line initialises the stack to 1 1 2 n, where n is the input number. The second line, running backwards, checks that n is greater than 1. If it is, n is decremented and the next element in the sequence is generated as follows:

r$:@+@r              a(n-3) a(n-2) a(n-1) n
                                           
r        Reverse   - n a(n-1) a(n-2) a(n-3)
 $       Swap      - n a(n-1) a(n-3) a(n-2)
  :      Duplicate - n a(n-1) a(n-3) a(n-2) a(n-2)
   @     Rotate 3  - n a(n-1) a(n-2) a(n-3) a(n-2)
    +    Add       - n a(n-1) a(n-2) a(n)
     @   Rotate 3  - n a(n) a(n-1) a(n-2)
      r  Reverse   - a(n-2) a(n-1) a(n) n

The final line outputs the number on the bottom of the stack, which is the required element in the sequence.

Python 2, 42 40 39 35 bytes

f=lambda n:n>1and f(n-2)+f(n-3)or 1

Generating the actual sets:

lambda n:["{:0{}b}".format(i,n).replace("0","-").replace("1","X")for i in range(2**n)if"11"not in"{:0{}b}".format(i*2,2+n).replace("000","11")]

J, 31 27 23 bytes

Saved 4 bytes thanks to miles!

0{]_&(]}.,+/@}:)1 1 2"_

An explanation is to come soon.

Old solution

(>.1&^)`(-&3+&$:-&2)@.(2&<)

This is an agenda. The LHS is a gerund composed of two verbs: >.1&^ and -&3+&$:-&2. The first one is used if the condition (2&<) fails. That means the fork >.1&^ is activated over the argument. Observe:

   1 ^ 0 1 2
1 1 1
   (1&^) 0 1 2
1 1 1
   0 1 2 >. (1&^) 0 1 2
1 1 2
   (>.1&^) 0 1 2
1 1 2

Here, >. takes the max of two values. Thus, it yields 1, 1, and 2 as the initial terms.

The second verb in the gerund is a fork:

-&3 +&$: -&2

The left and right tines are applied to the verb, subtracting 3 and 2 respectively; then the middle verb is called with left and right arguments equal to those. $: calls the verb on each argument, and + adds those two. It's basically equivalent to ($: arg - 3) + ($: arg - 2)

Test cases

   f =: (>.1&^)`(-&3+&$:-&2)@.(2&<)
   f 0
1
   f 2
2
   f 4
3
   f 6
5
   f 8
9
   F =: f"0         NB. for tables
   F i.13
1 1 2 2 3 4 5 7 9 12 16 21 28
   i.13
0 1 2 3 4 5 6 7 8 9 10 11 12
   (,. F) i.13
 0  1
 1  1
 2  2
 3  2
 4  3
 5  4
 6  5
 7  7
 8  9
 9 12
10 16
11 21
12 28

Jelly, 19 bytes

The recursive solution is probably shorter...

Ḥ⁹_c@⁸
+3µ:2R0;瀵S

See it at TryItOnline
Or see the series for n = [0, 99], also at TryItOnline

How?

Returns the n+3th Padovan's number by counting combinations

Ḥ⁹_c@⁸ - Link 1, binomial(k, n-2k): k, n
Ḥ      - double(2k)
 ⁹     - right argument (n)
  _    - subtract (n-2k)
     ⁸ - left argument (k)
   c@  - binomial with reversed operands (binomial(k, n-2k))

+3µ:2R0;瀵S - Main link: n
  µ       µ  - monadic chain separation
+3           - add 3 (n+3)
   :2        - integer divide by 2 ((n+3)//2)
     R       - range ([1,2,...,(n+3)//2]
      0;     - 0 concatenated with ([0,1,2,...,(n+3)//2]) - our ks
        ç€   - call previous link as a dyad for each
           S - sum

AnyDice, 51 bytes

function:A{ifA<3{result:(A+2)/2}result:[A-2]+[A-3]}

There should be more AnyDice answers around here.

My solution defines a recursive function that calculates a(n)=a(n-2)+a(n-3). It returns a(0)=a(1)=1 and a(2)=2 using some integer division magic.

Try it online

Note: the output may look weird, and that's because it's usually used to output dice probabilities. Just look at the number to the left of the bar chart.

Jelly, 11 bytes

,’fR_2߀So1

Try it online! or verify all test cases.

How it works

,’fR_2߀So1  Main link. Argument: n

 ’           Decrement; yield n - 1.
,            Pair; yield [n, n - 1].
   R         Range; yield [1, ..., n].
  f          Filter; keep the elements that are common to both lists.
             This yields [n, n - 1] if n > 1, [1] if n = 1, and [] if n < 1.
    _2       Subtract 2 from both elements, yielding [n - 2, n - 3], [-1], or [].
      ߀     Recursively call the main link for each integer in the list.
        S    Take the sum of the resulting return values.
         o1  Logical OR with 1; correct the result if n < 1.

Python, 33 bytes

f=lambda n:+(n<2)or f(n-2)+f(n-3)

Uses the shifted base cases f(-1) = f(0) = f(1) = 1. If True could be used for 1, we would not need 3 bytes for the +().

Mathematica, 35 bytes

a@0=a@1=1;a@2=2;a@b_:=a[b-2]+a[b-3]

Defines a function a. Takes an integer as input and returns an integer as output. Simple recursive solution.

JavaScript (ES6), 62 bytes

n=>[1,...Array(n)].reduce(($,_,i,a)=>a[i]=i<3?i:a[i-3]+a[i-2])

This is the first time that I've needed two dummy variable names. A recursive version would probably be shorter, but I really like reduce... Edit: Found a solution, also 62 bytes, which only has one dummy variable:

n=>[1,...Array(n)].reduce((p,_,i,a)=>a[i]=i<5?i+2>>1:a[i-5]+p)

Java 7, 65 42 bytes

int g(int u){return u>1?g(u-2)+g(u-3):1;}

The sequence just adds previous elements to get new ones. Hat tip to orlp and Rod for this shorter method ;)

Old:

int f(int u){return u<6?new int[]{1,1,2,2,3,4}[u]:f(u-1)+f(u-5);}

After the fifth element, the gap in the sequence rises by the element five previous.

PHP, 105 113 93 bytes

+3 for n=1; +9 for $argv, -1-3 golfed
-20: noticed that I don´t have to the combinations, but only their count

for($i=1<<$n=$argv[1];$i--;)$r+=!preg_match("#11|(0|^)0[0,]#",sprintf("%0{$n}b,",$i));echo$r;

run with -r

loop from 2**n-1 to 0:

• check binary n-digit representation for 11, 000, 00 at the beginning or the end, or a single 0
• if no match, increase result

print result

same size, slightly simpler regex

for($i=1<<$n=$argv[1];--$i;)$r+=!preg_match("#11|^00|00[,0]#",sprintf("%0{$n}b,",$i));echo$r;

• loop from 2**n-1 to 1 (instead of 0)
• check binary representation for 11, 00 at the beginning or the end, or 000
• prints nothing for n=0

PHP, 82 bytes

Arnauld´s answer ported and golfed:

for($i=$k=1<<$n=$argv[1];--$i;)$r+=!($i&$x=$i/2|$i*2)&&(($i|$x)&~$k)==$k-1;echo$r;

prints nothing for n=0

MATL, 25 23 bytes

W:qB7BZ+t!XAw3BZ+!3>a>s

Try it online! Or check all test cases.

Explanation

Two convolutions! Yay!

This builds an array, say A, where each possible configuration is a row. 1 in this array represents an occupied position. For example, for input 4 the array A is

0 0 0 0
0 0 0 1
0 0 1 0
···
1 1 1 0
1 1 1 1

The code then convolves array A with [1 1 1]. This gives an array B. Occupied positions and neighbours of occupied positions in A give a nonzero result in array B:

0 0 0 0
0 0 1 1
0 1 1 1
···
2 3 2 1
2 3 3 2

So the first condition for a configuration to be a checkmate is that B contains no zeros in that row. This means that in that row of A there no empty positions, or there were some but were neighbours of occupied positions.

We need a second condition. For example, the last row fulfills the above condition, but is not part of the solution because the configuration was not valid to begin with. A valid configurations cannot have two neighbouring occupied positions, i.e cannot have two contiguous 1's in A. Equivalently, it cannot have two contiguous values in B exceeding 1. So we can detect this by convolving B with [1 1] and checking that in the resulting array, C,

0 0 0 0
0 1 2 1
1 2 2 1
···
5 5 3 1
5 6 5 2

no value in that row exceeds 3. The final result is the number of configurations that fulfill the two conditions.

W:q    % Range [0 1 ... n-1], where n is implicit input
B      % Convert to binary. Each number produces a row. This is array A
7B     % Push array [1 1 1] 
Z+     % 2D convolution, keeping size. Entries that are 1 or are horizontal 
       % neighbours of 1 produce a positive value. This is array B
t!     % Duplicate and transpose (rows become columns)
XA     % True for columns that contain no zeros
w      % Swap. Brings array B to top
3B     % Push array [1 1]
Z+     % 2D convolution, keeping size. Two horizontally contiguous entries
       % that exceed 1 will give a result exeeding 3. This is array C
!      % Transpose
3>     % Detect entries that exceed 3
a      % True for columns that contain at least one value that exceeds 3
>      % Element-wise greater-than comparison (logical and of first
       % condition and negated second condition)
s      % Sum (number of true values)

Ruby, 58 34 bytes

Heavily inspired by Geobits' original Java answer.

f=->n{n<3?n:n<6?n-1:f[n-1]+f[n-5]}

See it on repl.it: https://repl.it/Dedh/1

First attempt

->n{(1...2**n).count{|i|!("%0#{n}b"%i)[/11|^00|000|00$/]}}

See it on repl.it: https://repl.it/Dedh

Oasis, 5 bytes

Code

cd+2V

Extended version

cd+211

Explanation

1 = a(0)
1 = a(1)
2 = a(2)

a(n) = cd+
       c      # Calculate a(n - 2)
        d     # Calculate a(n - 3)
         +    # Add them up

Try it online!