A213727 a(n) = 0 if n is in the infinite trunk of the "binary beanstalk", otherwise number of nodes (including leaves and the node n itself) in that finite branch of the beanstalk.

0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 3, 0, 1, 1, 1, 0, 0, 1, 3, 0, 1, 1, 5, 0, 1, 3, 0, 1, 1, 1, 1, 0, 0, 1, 3, 0, 1, 1, 5, 0, 1, 3, 0, 1, 1, 1, 0, 7, 1, 0, 5, 1, 1, 0, 3, 1, 3, 0, 1, 1, 1, 1, 1, 0, 0, 1, 3, 0, 1, 1, 5, 0, 1, 3, 0, 1, 1, 1, 0, 7, 1, 0, 5, 1, 1, 0, 3

Offset: 0

Author

Antti Karttunen, Nov 01 2012

Keywords

• nonn

Comments

a(n) tells, for each natural number n, whether it belongs to the infinite trunk of the binary beanstalk (in A179016, when a(n)=0), or if it is one of the terminal nodes (i.e., leaves, A055938, when a(n)=1), or otherwise, when a(n) > 1, tells from how many different numbers one can arrive at this n by repeatedly subtracting their bit-count (A000120) from them (as explained in A071542), and including also n itself as one of the cases.

a(n) is always odd unless it is zero. In particular, each a(A213717(n)) is an odd number greater than one.

Examples

			a(10) = 3 because we include 10 itself ("1010" in binary) and the two numbers n for which it is true that n - A000120(n) = 10, i.e., 12 and 13 ("1100" and "1101" in binary). Furthermore, there do not exist any such numbers for 12 or 13, as both are members of A055938 (see also the comment at A213717).
Similarly, a(22) = 5 as there are the following five cases: 22 itself, 24 as 24-A000120(24) = 24-2 = 22 (note that 24 is in A055938), 25 as 25-A000120(25) = 25-3 = 22, and the two terminal nodes (leaves) branching from 25, that is, 28 & 29 (as 28-A000120(28) = 28-3 = 25, and 29-A000120(29) = 29-4=25).
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..16384

Crossrefs

Cf. A179016.

Cf. also A055938, A079559, A213717, A213719, A213723, A213724, A213725, A213726, A213731.

Analogous sequences computed for other "beanstalk" or similar tree systems: A227643, A230427, A255327.

Formula

If A079559(n)=0, a(n)=1; otherwise, if A213719(n)=1, a(n)=0; otherwise a(n) = 1+a(A213723(n))+a(A213724(n)).

Other identities. For all n:

a(A179016(n)) = 0, a(A055938(n)) = 1, and a(A213717(n)) >= 3.

a(A213717(n)) = (2*A213726(A213717(n)))-1.