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, 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

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 A344593 A353278

Adjacent sequences: A306919 A306920 A306921 * A306923 A306924 A306925

Keyword

nonn,base

Author

Peter Kagey, Mar 16 2019

Status

approved