Source code in the clu programming language

figfig = cluster is ffr, ffs
    rep = null
    ai = array[int]
    own R: ai := ai$[1]
    own S: ai := ai$[2]
    
    % Extend R and S until R(n) is known
    extend = proc (n: int)
        while n > ai$high(R) do
            next: int := ai$top(R) + S[ai$high(R)]
            ai$addh(R, next)
            while ai$top(S) < next-1 do
                ai$addh(S, ai$top(S)+1)
            end
            ai$addh(S, next+1)
        end
    end extend
    
    ffr = proc (n: int) returns (int)
        extend(n)
        return(R[n])
    end ffr
    
    ffs = proc (n: int) returns (int)
        while n > ai$high(S) do
            extend(ai$high(R) + 1)
        end
        return(S[n])
    end ffs
end figfig

start_up = proc ()
    ai = array[int]
    po: stream := stream$primary_output()
    
    % Print R[1..10]
    stream$puts(po, "R[1..10] =")
    for i: int in int$from_to(1,10) do
        stream$puts(po, " " || int$unparse(figfig$ffr(i)))
    end
    stream$putl(po, "")
    
    % Count the occurrences of 1..1000 in R[1..40] and S[1..960]
    occur: ai := ai$fill(1, 1000, 0)
    for i: int in int$from_to(1, 40) do
        occur[figfig$ffr(i)] := occur[figfig$ffr(i)] + 1
    end
    for i: int in int$from_to(1, 960) do
        occur[figfig$ffs(i)] := occur[figfig$ffs(i)] + 1
    end
    
    % See if they all occur exactly once
    begin
        for i: int in int$from_to(1, 1000) do
            if occur[i] ~= 1 then exit wrong(i) end
        end
        stream$putl(po, 
            "All numbers 1..1000 occur exactly once in R[1..40] U S[1..960].")
    end except when wrong(i: int):
        stream$putl(po, "Error: " ||
            int$unparse(i) || " occurs " || int$unparse(occur[i]) || " times.")
    end
end start_up