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The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.

From Antti Karttunen, Jun 27 2014: (Start)

The odd bisection (containing even terms) halved gives A244153.

The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.

(End)

Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

This sequence can be used for filtering certain base-2 related sequences, because it matches only with any such sequence b that can be computed as b(n) = f(A005940(n+1)), where f(n) is any function that depends only on the prime signature of n (some of these are listed under the index entry for "sequences computed from exponents in ...").

Matching in this context means that the sequence a matches with the sequence b iff for all i, j: a(i) = a(j) => b(i) = b(j). In other words, iff the sequence b partitions the natural numbers to the same or coarser equivalence classes (as/than the sequence a) by the distinct values it obtains.

Because the Doudna map n -> A005940(1+n) is an isomorphism from "unary-binary encoding of factorization" (see A156552) to the ordinary representation of the prime factorization of n, it follows that the equivalence classes of this sequence match with any such sequence b, where b(n) is computed from the lengths of 1-runs in the binary representation of n and the order of those 1-runs does not matter. Particularly, this holds for any sequence that is obtained as a "Run Length Transform", i.e., where b(n) = Product S(i), for some function S, where i runs through the lengths of runs of 1's in the binary expansion of n. See for example A227349.

However, this sequence itself is not a run length transform of any sequence (which can be seen for example from the fact that A046523 is not multiplicative).

Furthermore, this matches not only with sequences involving products of S(i), but with any sequence obtained with any commutative function applied cumulatively, like e.g., A000120 (binary weight, obtained in this case as Sum identity(i)), and A069010 (number of runs of 1's in binary representation of n, obtained as Sum signum(i)).

a(n) can be also computed by replacing all consecutive runs of zeros in the binary expansion of n with * (multiplication sign), and then performing that multiplication, still in binary, after which the result is converted into decimal. See the example below.

For the definition of the Run Length Transform see A246595.

This is the Run Length Transform of S(n) = Fibonacci(n+2).

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

Also the number of sparse subsets of the binary indices of n, where a set is sparse iff 1 is not a first difference. The maximal case is A384883. For prime instead of binary indices we have A166469. - Gus Wiseman, Jul 05 2025

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

This sequence is obtained by applying Run Length Transform to the right-shifted version of the sequence A001317: 1, 3, 5, 15, 17, 51, 85, 255, 257, ...

a(0) = 0 (an empty sum).

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

This sequence is obtained by applying Run Length Transform to sequence b = 1, 3, 5, 17, 257, 65537, ... (1 followed by Fermat numbers, with b(1) = 1, b(2) = 3, b(3) = 5, ..., b(n) = 2^(2^(n-2)) + 1 for n >= 2).