

A088748


a(n) = 1 + Sum_{k=0..n1} 2 * A014577(k)  1.


5



1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 5, 6, 7, 6, 5, 6, 5, 4, 3, 4, 5, 4, 5, 6, 5, 4, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 5, 6, 7, 6, 5, 6, 5, 4, 5, 6, 7, 6, 7, 8, 7, 6, 5, 6, 7, 6, 5, 6, 5, 4, 3, 4, 5, 4, 5, 6
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OFFSET

0,2


COMMENTS

Let s(0)=1; s(n+1)=s(n),ri(n), where ri(n) is s(n) reversed and incremented. Each s(n) is an initial part of this sequence.
For each m, a(1 to 2^m) is a permutation of A063787(1 to 2^m). For k=1 to 2^m, a(2^m+1A088372(m,k)) = A063787(k).
Partial sums give A164910: (1, 3, 6, 8, 11, 15, 20, ...).
a(0) = 1, then using the dragon curve sequence A014577: (1, 1, 0, 1, 1, ...) as a code: (1 = add to current term, 0 = subtract from current term, to get the next term), see example.
Rows of A088696 tend to this sequence.


LINKS



FORMULA



EXAMPLE

The first 8 terms of the sequence = (1, 2, 3, 2, 3, 4, 3, 2), where the first four terms = (1, 2, 3, 2). Reverse, add 1, getting (3, 4, 3, 2), then append.
The sequence begins with "1", then using the dragon curve coding, we get:
....1...1...0...1...1... = A014577, the dragon curve.


MATHEMATICA

Array[1 + Sum[2 (1  (((Mod[#1, 2^(#2 + 2)]/2^#2))  1)/2)  1 &[k, IntegerExponent[k, 2]], {k, #  1}] &, 102] (* Michael De Vlieger, Aug 26 2020 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



