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

base = 2

\\ can the digits of x and of y be combined to form a palindrome ?
ok(x,y) = {
	my (d = concat(apply(v -> if (v==0, [0], digits(v, base)), [x,y])), f=vector(base));
	for (i=1, #d, f[d[i]+1]++);
	#select(v -> v%2, f)<=1 && f[1]+1!=#d
}

\\ least v not yet in the sequence whose digits can be combined with those of p to form a palindrome
other(p) = {
	see(p);
	for (v=unseen, oo,
		if (!seen(v) && ok(p,v),
			return (v);
		);
	);
}

\\ combine digits of x and y to form the least palindrome
mix(x,y) = {
	my (d = concat(apply(v -> if (v==0, [0], digits(v, base)), [x,y])), f=vector(base));
    for (i=1, #d, f[d[i]+1]++);
	my (m=0);	\\ middle digit
	if (#d%2,
		for (i=1, base,
			if (f[i]%2,
				m = i-1;
				f[i]--;
				break;
			);
		);
	);
	my (left=0);
	for (i=2, base,
		if (f[i],
			left = i-1;
			f[i]-=2;
			break;
		);
	);
	for (i=1, base,
		while (f[i],
			left = base*left + i-1;
			f[i]-=2;
		);
	);

	return ((left * base^(#d%2) + m) * base^(#d\2) + fromdigits(Vecrev(digits(left, base)), base));
}

my (p=0); for (n=0, 10 000, v=other(p); print (n " " fromdigits(digits(mix(p,v), base))); p=v);

quit