s = 0
unseen = 1
seen(v) = bittest(s, v)
see(v) = s = bitor(s, 2^v); while (seen(unseen), unseen++)

P = 3	\\ P = 2 for A005941 	Inverse of the Doudna sequence A005940.

a = vector(10 000)	\\ inverse of A356867
u = 1

seq = [0]	\\ [n, a(n), p]
nb = 0

emit(n, v) = {
	if (v <= #a,
		a[v] = n;
		while (a[u],
			print (u " " a[u]);
			if (u++ > #a,
				quit;
			);
		);

		see(v);

		if (#seq < nb++,
			seq = concat(seq, vector(#seq));
		);

		seq[nb] = [n, v, 2];
	);
}

{
	for (e = 0, oo,
		my (t = P^e, v);
		emit(t, t);
		for (m = 1, oo,
			if (m > nb,
					break,

				seq[m][1] >= (P-1)*t,
					break,

					while (seq[m][3]==P || seen(v = seq[m][2] * seq[m][3]),
						seq[m][3] = nextprime(1 + seq[m][3]);
					);
					emit(t + seq[m][1], v);
			);
		);
	);
}