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Allowing for offset, also run lengths in the parity of A171791.

Records appear to be given by A001045 Jacobsthal numbers.

Run lengths of repeated 1's = 5,3,5,5,3,5,3,5,5,3,5, appears to be A194584.

Number of terms in interval [2^(n-1), 2^n) is the number of compositions of n with distinct parts (cf. A032020). For example, if n=6, then interval [2^5, 2^6) contains 11 terms {32,...,52}. This corresponds to 11 compositions with distinct parts of 6: 6, 5+1, 1+5, 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3.

From Gus Wiseman, Apr 06 2020: (Start)

The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. This sequence lists all numbers k such that the k-th composition in standard order is strict. For example, the sequence together with the corresponding strict compositions begins:

0: () 38: (3,1,2) 98: (1,4,2)

1: (1) 40: (2,4) 104: (1,2,4)

2: (2) 41: (2,3,1) 128: (8)

4: (3) 44: (2,1,3) 129: (7,1)

5: (2,1) 48: (1,5) 130: (6,2)

6: (1,2) 50: (1,3,2) 132: (5,3)

8: (4) 52: (1,2,3) 133: (5,2,1)

9: (3,1) 64: (7) 134: (5,1,2)

12: (1,3) 65: (6,1) 137: (4,3,1)

16: (5) 66: (5,2) 140: (4,1,3)

17: (4,1) 68: (4,3) 144: (3,5)

18: (3,2) 69: (4,2,1) 145: (3,4,1)

20: (2,3) 70: (4,1,2) 152: (3,1,4)

24: (1,4) 72: (3,4) 160: (2,6)

32: (6) 80: (2,5) 161: (2,5,1)

33: (5,1) 81: (2,4,1) 176: (2,1,5)

34: (4,2) 88: (2,1,4) 192: (1,7)

37: (3,2,1) 96: (1,6) 194: (1,5,2)

(End)

The standard order of compositions is given by A066099.

From Vladimir Shevelev, Dec 18 2013: (Start)

Every number in binary is a concatenation of parts of the form 10...0 with k>=0 zeros. For example, 5=(10)(1), 11=(10)(1)(1), 7=(1)(1)(1). We call d>0 a c-divisor of m, if d consists of some consecutive parts of m taking from the left to the right. Note that, to d=0 corresponds an empty set of parts. So it is natural to consider 0 as a c-divisor of every m. For example, 5=(10)(1) is a divisor of 23=(10)(1)(1)(1). Analogously, 1,2,3,7,11,23 are c-divisors of 23. But 6=(1)(10) is not a c-divisor of 23.

One can prove a one-to-one correspondence between distinct subsequences for composition no. n in standard order and c-divisors of n. So, the sequence lists also numbers of c-divisors of nonnegative integers.

(End)

These are contiguous subsequences, or restrictions to a subinterval. The case for all subsequences is A334299. - Gus Wiseman, Jun 02 2020

From Vladimir Shevelev, Dec 11 2014: (Start)

Or, sum of parts of the form 10...0 with nonnegative number of zeros in binary representation of n as the corresponding powers of 2. For example, n=50 in binary is a concatenation of parts (1)(100)(10). Then a(50)=1+4+2=7.

Every positive number k occurs a finite number of times, such that the position of the last appearance of k is 2^k-1.

Moreover, the number of times of appearances of k is the number of compositions of k into powers of 2, i.e., it is A023359(k), k>0. (End)

Conjecture: the sequence is bounded by a constant.

The number of appearances of k is the number of compositions of k into numbers of the form 3^n and 2*3^n, A235684(k).

For c-equivalence, see the comments to A233249. Briefly put, two positive integers m and n are c-equivalent in the sense of Vladimir Shevelev, if they have ordinary binary representations with the same multisets of substrings resulting from cutting the full strings immediately before each bit 1. A(n) is defined as the smallest positive integer, which is c-equivalent to n. Alternatively, in the manners of A114994, the lengths of these substrings can be considered as representing ways to write integers as sums of positive integers with arbitrarily ordered sums, and a(n) as the unique integer whose corresponding substring lengths form the corresponding integer partition.

For instance, the ordinary binary representations of 11, 13, and 14 are 1011, 1101, and 1110, respectively, which yields the equal multisets {"10","1","1"}, and {"1","10","1"}, and {"1","1","10"} of strings, respectively; whence 11, 13, and 14 are c-equivalent.

A(n) is an odd number if and only if the substring "1" appears at least once in the multiset. Since this is the case if and only if it also holds for the binary representation of n concatenated with itself, and A233312(n) = a(m) for the number m whose binary representation is this concatenation, we have a(n) == A233312(n) (mod 2) for all n. Moreover, empirical data has suggested that perhaps A233312(n)+1 == A171791(n+1) (mod 2) for all n >= 1. This relation holds in general if and only if a(n)+1 == A171791(n+1) for the same n, which in its turn is true if and only if the relation with fibbinary numbers first empirically observed by Paul D. Hanna in the comments to A171791 holds in general.

The sequence A163382 also maps n to a c-equivalent integer <=n; however, here, only cyclic permutations of the sequences of substrings are allowed. Thus, a more restricted equivalence relation is used; whence a(n) <= A163382(n) for all n. Equality holds for infinitely many n, including n = 1..37.