

A306922


Number of distinct powers of two obtained by breaking the binary representation of n into consecutive blocks, and then adding the numbers represented by the blocks.


1



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

1,2


COMMENTS

1's appear at indices given by A321321.


LINKS

Peter Kagey, Table of n, a(n) for n = 1..10000
Elwyn Berlekamp and Joe P. Buhler, Puzzle 6, Puzzles column, Emissary, MSRI Newsletter, Fall 2011, Page 9, Problem 6.
Reddit user HarryPotter5777, Partition a binary string so sum of chunks is a power of two. (Proposed proof that a(n) > 0 for all n.)


EXAMPLE

For n = 46, the a(46) = 3 powers of two that come from the partition of "101110" are 4, 8, and 16:
[10, 1110] > [2, 14] > 16
[1, 0, 1, 110] > [1, 0, 1, 6] > 8
[101, 1, 10] > [5, 1, 2] > 8
[1, 0, 111, 0] > [1, 0, 7, 0] > 8
[101, 11, 0] > [5, 3, 0] > 8
[1, 0, 1, 1, 1, 0] > [1, 0, 1, 1, 1, 0] > 4


CROSSREFS

Cf. A306921, A321318, A321319, A321320, A321321.
Sequence in context: A326674 A331184 A333768 * A331181 A328917 A265917
Adjacent sequences: A306919 A306920 A306921 * A306923 A306924 A306925


KEYWORD

nonn,base


AUTHOR

Peter Kagey, Mar 16 2019


STATUS

approved



