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When filtering sequences (by equivalence class partitioning), this sequence can be used instead of A278222, because for all i, j it holds that: a(i) = a(j) <=> A278222(i) = A278222(j).

For example, for all i, j: a(i) = a(j) => A000120(i) = A000120(j), and for all i, j: a(i) = a(j) => A001316(i) = A001316(j).

The sequence allots a distinct value for each distinct multiset formed from the lengths of 1-runs in the binary representation of n. See the examples. - Antti Karttunen, Jun 04 2017

Term a(n) essentially records the run lengths of numbers of form 4k+3 encountered when starting from that node in binary tree A005940 which contains n, and by then traversing towards the root by iterating the map n -> A252463(n). The actual run lengths can be read from the exponents of primes in the prime factorization of A278222(A292383(m)), where m = min_{k=1..n} for which a(k) = a(n). In compound filter A292584 this is combined with similar information about the run lengths of the numbers of the form 4k+1 (A292585).

From Antti Karttunen, Sep 25 2017: (Start)

For all i, j: a(i) = a(j) => A053866(i) = A053866(j).

This follows from the interpretation of A053866 (A093709) as the characteristic function of squares and twice-squares. In binary tree A005940 each number is "born" by repeated applications of two functions: when we descend leftward we apply A003961, which shifts all prime factors of n one step towards larger primes. On the other hand, when we descend rightward the terms grow by doubling: n -> 2n (A005843). No square is ever of the form 4k+3, and for any square x, A003961(x) is also a square. Multiplying a square by 2 gives twice a square, and then multiplying by 2 again gives 4*square, which is also a square. In general, applying an even number of doubling steps in succession keeps a square as a square, while an odd number of doubling steps gives twice a square. Applying A003961 to any 2*square gives 3*(some square) which is always of the form 4k+3. Moreover, after any such "wrong turn" in A005940-tree no square nor twice a square can ever be encountered under any of the further descendants, because with this process it is impossible to find a pair for the lone prime factor now present. On the other hand, when turning left from any (2^2k)*s (where s is a square), one again gets a square of the form (3^2k)*A003961(s). All this implies that there are no numbers of the form 4k+3 in any trajectory leading to a square or twice a square in A005940-tree, while all trajectories to any other kind of number contain at least one number of the form 4k+3. Because each a(n) in this sequence contains enough information to count the 4k+3 numbers encountered on a A005940-trajectory to n (being 1 iff there are none), this filter matches A053866.

(End)

Positions of 2's is given by the positive Fibonacci numbers: 1, 2, 3, 5, 8, 13, 21, ..., that is, A000045(n) from n >= 2 onward.

Positions of 3's is given by Lucas numbers larger than 3: 4, 7, 11, 18, ..., that is, A000032(n) from n >= 3 onward.

Sequence allots a distinct value for each distinct multiset formed from the lengths of 1-runs in the binary representation of A048679(n). Compare to the scatter plot of A286622.

Restricted growth sequence transform of the ordered pair [A278222(n), A336158(n)], i.e., of the ordered pair [A046523(A005940(1+n)), A046523(A000265(n))].

For all i, j: A324400(i) = A324400(j) => A003602(i) = A003602(j) => a(i) = a(j).

For all i, j: a(i) = a(j) => A304536(i) = A304536(j).

Restricted growth sequence transform of the ordered pair [A278222(n), A324342(n)], or equally, of [A286622(n), A324342(n)].

For all i, j: a(i) = a(j) => A324344(i) = A324344(j).