Javascript 117 111 bytes

Thanks to @theonlygusti for helping golf off 5 bytes

x=>{r=0;a=[1,2,3];i=1;while(a[++i]<x)a[i+1]=a[i]+a[i-1]+a[i-2];for(;x;i--)if(x>=a[i]){r+=1<<i;x-=a[i]}return r}

How It Works

First, the function generates all tribonacci numbers until it finds one greater than the input

a=[1,2,3];i=1;for(;a[++i]<x;)a[i+1]=a[i]+a[i-1]+a[i-2];

Next, it reverse searches the list of numbers. If a number is less than the input, it adds 2^(index of that number) to the return value and reduces the input by that number.

for(;x;i--)if(x>=a[i]){r+=1<<i;x-=a[i]}

Finally it returns the result.

Try it Online

Python 2, 110 102 bytes

-3 bytes thanks to Rod (neat trick for casting boolean i to an int with +i so the repr `+i` works)

n=input()
x=[3,2,1]
r=''
while x[0]<n:x=[sum(x[:3])]+x
for v in x:i=n>=v;n-=v*i;r+=`+i`
print int(r,2)

Try it online!

JavaScript (ES6), 97 93 bytes

Here, we are using reduce() on a recursive function. We assume that the output is 31-bit (which is the largest unsigned quantity that JS can easily work with for bitwise operations anyway).

n=>[...Array(x=31)].reduce(p=>(c=(F=k=>k<4?k:F(--k)+F(--k)+F(--k))(x--))>n?p:(n-=c,p|1<<x),0)

Performance wise, this is clearly not very efficient.

For the curious:

• The ratio between the number of calls to F() for N+1 reduce() iterations vs N iterations quickly converges towards the Tribonacci Constant (≈ 1.83929). Therefore, each additional bit in the output costs approximately twice as much time as the previous one.
• With 31 bits, the F() function is called a good 124 million times.

Test

NB: This may take 1 or 2 seconds to complete.

let f =

n=>[...Array(x=31)].reduce(p=>(c=(F=k=>k<4?k:F(--k)+F(--k)+F(--k))(x--))>n?p:(n-=c,p|1<<x),0)

console.log(f(63))

Mathematica, 78 74 bytes

Fold[#+##&,#~NumberDecompose~Reverse@LinearRecurrence[{1,1,1},{1,2,3},#]]&

LinearRecurrence[{1,1,1},{1,2,3},#] generates a list, of length equal to the input, of the 1-2-3 tribonacci numbers. (The {1,1,1} represents the sum of the previous three terms, while {1,2,3} are the initial values.) Then #~NumberDecompose~ finds the greediest way to write the input as a sum of elements of the list (this is the same function that would decompose a monetary amount into multiples of the available currencies, for example). Finally, Fold[#+##&,...] converts the resulting binary list into a (base-10) integer.

Previous submission:

Fold[#+##&,#~NumberDecompose~Reverse@Array[If[#<4,#,Tr[#0/@(#-Range@3)]]&,#]]&

As is often the case (though not above), this golfed version is super slow on inputs larger than 20 or so, because it generates (with non-optimized recursion) a list of tribs whose length is the input; replacing the final # by a more reasonable bound like Round[2Log@#+1] results in much better performance.

Haskell, 95 bytes

(a!b)c=a:(b!c)(a+b+c)
e#(r,c)|c-e<0=(2*r,c)|1<2=(2*r+1,c-e)
f n=fst$foldr(#)(0,n)$take n$(1!2)3

Usage example: f 63 -> 104. Try it online!.

How it works: ! builds the 1-2-3-Tribonacci sequence. Given 1, 2 and 3 as the start parameters, we take the first n elements of the sequence. Then we fold from the right function # which subtracts the next element e from n and sets the bit in the return value r if e is needed or lets the bit unset. Setting the bit is doubling r and adding 1, letting it unset is just doubling.

Jelly, 31 bytes

S=³
3RUµḣ3S;µ<³Ạµ¿µŒPÇÐfṀe@ЀµḄ

Try it online!

I'm almost certain there is a MUCH shorter way to achieve this in Jelly.

How?

S=³ - Link 1, compare sum to input: list
S   - sum
  ³ - 3rd command line argument which is 1st program argument.
 =  - equal?

3RUµḣ3S;µ<³Ạµ¿µŒPÇÐfṀe@ЀµḄ - Main link: n
3RU                         - range(3) upended -> [3,2,1]
   µ    µ   µ¿              - while
         <³                 - less than input (vectorises)
           Ạ                - all?
    ḣ3S;                    -     head(3), sum, and concatenate
                                  [3,2,1] -> [6,3,2,1] -> [11,6,3,2,1] -> ...
              µ             - monadic chain separation, call the result x
               ŒP           - power set of x - e.g. for [6,3,2,1] -> [[],[6],[3],[2],[1],[6,3],[6,2],[6,1],[3,2],[3,1],[2,1],[6,3,2],[6,3,1],[6,2,1],[3,2,1],[6,3,2,1]]
                  Ðf        - filter keep
                 Ç          -     last link (1) as a monad (those that sum to the input)
                    Ṁ       - maximum (e.g. an input of 63 would yield [[37,20,6],[37,20,3,2,1]], the maximum of which is [37,20,6], the one with the largest numbers used)
                         µ  - monadic chain separation (to have x as the right argument below)
                     e@Ѐ   - exists in with reversed arguments mapped over x (e.g. [37,20,6] with x = [68,37,20,11,6,3,2,1] yields [0,1,1,0,1,0,0,0])
                          Ḅ - convert from binary to integer.        

Perl 6, 93 91 bytes

-2 bytes thanks to b2gills

{my@f=1,2,3,*+*+*...*>$^n;sum @f».&{$_~~any first *.sum==$n,@f.combinations}Z*(2 X**^∞)}

How it works

• First, it generates the 1-2-3-Tribonacci sequence up to the first element larger than the input:

  my @f = 1, 2, 3, *+*+* ... * > $^n;

• Based on that it finds the subset of the sequence which adds up to the input:

  first *.sum==$n, @f.combinations

• Based on that it constructs a list of booleans specifying whether each element of the sequence is part of the sum:

  @f».&{$_~~any ...}

• And finally it interprets that list of True=1, False=0 values as base 2 and returns it as a (base 10) number:

  sum ... Z* (2 X** ^∞)

JavaScript (ES6), 61 60 bytes

n=>(g=(x,y,z)=>(n>x&&g(y,z,x+y+z)*2)+!(n<x||![n-=x]))(1,2,3)

Calculates the 1-2-3-Tribonacci numbers until it reaches the original number, then as the recursion unwinds, tries to subtract each one in turn, doubling the result as it goes.

Edit: Saved 1 byte thanks to @Arnauld.

Batch, 151 148 145 bytes

@set/ar=0,n=%1
@call:c 3 2 1
@echo %r%
@exit/b
:c
@set/as=%1+%2+%3
@if %n% gtr %3 call:c %s% %*
@set/ar*=2
@if %n% geq %3 set/an-=%3,r+=1

Port of my JavaScript answer. Edit: Saved 3 bytes by passing my subroutine arguments in reverse order and another 3 bytes by using individual @s on each line instead of @echo off.

Jelly (fork), 17 bytes

3RṚµḣ3S;µ³¡
æF¢ṪḄ

This relies on a fork of Jelly where I am disappointingly still working on implementing an efficient Frobenius solve atom. For those who are interested, I would like to match Mathematica's speed in FrobeniusSolve and luckily there is an explanation of their method in the paper "Making Change and Finding Repfigits: Balancing a Knapsack" by Daniel Lichtblau.

Explanation

3RṚµḣ3S;µ³¡  Helper link. Nilad - no arguments
3            The constant 3
 R           Range. Makes [1, 2, 3]
  Ṛ          Reverse. Makes [3, 2, 1]
   µ         Begin new monadic chain operating on [3, 2, 1]
        µ    Create monadic chain
    ḣ3         Take the first 3 items
      S        Sum
       ;       Prepend to the list
         ³¡  Repeat it n (initial argument from main) times

æF¢ṪḄ        Main link. Input: integer n
  ¢          Call helper link as a nilad
æF           Frobenius solve for n using the first n 123-Tribonacci values in reverse
   Ṫ         Tail. Take the last value. The results of Frobenius solve are ordered
             where the last result uses the least
    Ḅ        Unbinary. Convert digits from base 2 to base 10