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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)

Term a(n) essentially records the run lengths of numbers of form 4k+1 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(A292385(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+3 (A292583).

This is essentially also a filter constructed from the runlengths of numbers of the form 4k+0 and the runlengths of numbers of the form 4k+2 encountered in trajectories of A005940-tree. See comments in A083399 and A292586.

For all i, j: A291757(i) = A291757(j) => a(i) = a(j), that is, this filter matches to a subset of the sequences matched by filter A291757.

Moreover, for all i, j: a(i) = a(j) <=> A101296(i) = A101296(j), thus the subset is exactly the sequences matched by A101296 (A046523). This follows because the prime signature of n can be recovered from the two components as A046523(n) = A046523(A003557(n)) * A292586(n) and also vice versa as A046523(A003557(n)) = A003557(A046523(n)).

A filter constructed from the runlengths of numbers of the form 4k+0 and the runlengths of numbers of the form 4k+3 encountered in trajectories of A005940-tree.

Restricted growth sequence transform of the ordered pair [A336312(n), A336313(n)].

For all i, j:

A305800(i) = A305800(j) => a(i) = a(j) => A001222(i) = A001222(j).