s = 0
unseen = 1
seen(v) = bit test(s, v)
see(v) = s = bit or(s, 2^v); while (seen(unseen), unseen++)

a = vector(10 000 + inc=20)
open = 0	\\ mask of current open pairs

{
	for (n=1, #a-inc,
		if (a[n]==0,
			for (v=unseen, oo,
				if (!seen(v),
					my (h=digits(v)[1]);
					if (!bittest(open, h) && a[n+1+h]==0,
						see(a[n] = v);

						\\ search second term
						for (w=v+1, oo,
							if (!seen(w) && digits(w)[1]==h,
								see(a[n+1+h] = w);
								break;
							);
						);

						open += 2^h;
						break;
					);
				);
			),

			open -= 2^digits(a[n])[1]
		);
		print (n " " a[n])
	);
}

quit