A321320 -id:A321320 - cp's OEIS frontend

A321318 Number of distinct values obtained by partitioning the binary representation of n into consecutive blocks, and then summing the numbers represented by the blocks.

1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 8, 8, 7, 7, 9, 9, 8, 8, 7, 7, 6, 6, 9, 9, 9, 9, 10, 10, 9, 9, 9, 9, 12, 12, 13, 13, 9, 9, 12, 12, 12, 12, 12, 12, 12, 12, 14, 14, 12, 12, 11, 11, 7, 7, 11, 11, 12, 12, 13, 13, 12, 12, 15, 15, 15

Offset: 1

Jeffrey Shallit, Nov 04 2018

			For n = 13, we can partition its binary representation as follows (showing partition and sum of terms): (1101):13, (1)(101):6, (11)(01):4, (110)(1):7, (1)(1)(01):3, (1)(10)(1):4, (11)(0)(1):4, (1)(1)(0)(1):3.  Thus there are 5 distinct values.

Rémy Sigrist, Table of n, a(n) for n = 1..16384
Elwyn Berlekamp and Joe P. Buhler, Puzzle 6, Puzzles column, Emissary, MSRI Newsletter, Fall 2011, Page 9, Problem 6.
Steve Butler, Ron Graham, and Richard Stong, Collapsing numbers in bases 2, 3, and beyond, in The Proceedings of the Gathering for Gardner 10 (2012).
Steve Butler, Ron Graham, and Richard Strong, Inserting plus signs and adding, Amer. Math. Monthly 123 (3) (2016), 274-279.
Rémy Sigrist, PARI program for A321318

Cf. A321319, A321320, A321321.

PARI
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A321321 Numbers n for which the "partition-and-add" operation applied to the binary representation of n results in only one power of 2.

Original entry on oeis.org

1, 3, 5, 6, 7, 9, 11, 12, 13, 14, 17, 19, 21, 24, 25, 28, 31, 33, 35, 37, 41, 42, 48, 49, 56, 65, 67, 69, 73, 81, 87, 96, 97, 112, 129, 131, 133, 137, 145, 161, 167, 192, 193, 224, 257, 259, 261, 265, 273, 289, 321, 384, 385, 448, 513, 515, 517, 521, 529, 545

Offset: 1

Jeffrey Shallit, Nov 04 2018

nonn

Conjecture: With the exception of a(1) = 1 and a(17) = 31, all terms have a binary weight of 2 or 3. - Peter Kagey, Jun 14 2019

			For n = 13, we can partition its binary representation as follows (showing partition and sum of terms): (1101):13, (1)(101):6, (11)(01):4, (110)(1):7, (1)(1)(01):3, (1)(10)(1):4, (11)(0)(1):4, (1)(1)(0)(1):3.  Thus there is only one possible power of 2, namely 4.

Peter Kagey, Table of n, a(n) for n = 1..200
E. Berlekamp, J. Buhler, Puzzle 6, Puzzles column, Emissary Fall (2011) 9.
Steve Butler, Ron Graham, and Richard Stong, Collapsing numbers in bases 2, 3, and beyond, in The Proceedings of the Gathering for Gardner 10 (2012).
Steve Butler, Ron Graham, and Richard Strong, Inserting plus signs and adding, Amer. Math. Monthly 123 (3) (2016), 274-279.

Cf. A321318, A321319, A321320.

A321319 Smallest power of 2 obtainable by partitioning the binary representation of n into consecutive blocks and then summing.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 2, 4, 4, 4, 1, 2, 2, 4, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 4, 16, 1, 2, 2, 4, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 4, 8, 2, 4, 4, 4, 4, 4, 4, 8, 4, 4, 4, 8, 4, 8, 8, 8, 1, 2, 2, 4, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 4, 8, 2, 4, 4, 4, 4, 4, 4

Offset: 1

Jeffrey Shallit, Nov 04 2018

nonn

			For n = 13, we can partition its binary representation as follows (showing partition and sum of terms): (1101):13, (1)(101):6, (11)(01):4, (110)(1):7, (1)(1)(01):3, (1)(10)(1):4, (11)(0)(1):4, (1)(1)(0)(1):3.  Thus the smallest power of 2 is 4.

E. Berlekamp, J. Buhler, Puzzle 6, Puzzles column, Emissary Fall (2011) 9.
Steve Butler, Ron Graham, and Richard Stong, Collapsing numbers in bases 2, 3, and beyond, in The Proceedings of the Gathering for Gardner 10 (2012).
Steve Butler, Ron Graham, and Richard Strong, Inserting plus signs and adding, Amer. Math. Monthly 123 (3) (2016), 274-279.

Cf. A321318, A321320, A321321.

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.

Original entry on oeis.org

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

Peter Kagey, Mar 16 2019

1's appear at indices given by A321321.

			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

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.)

Cf. A306921, A321318, A321319, A321320, A321321.

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A321318 Number of distinct values obtained by partitioning the binary representation of n into consecutive blocks, and then summing the numbers represented by the blocks.

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A321321 Numbers n for which the "partition-and-add" operation applied to the binary representation of n results in only one power of 2.

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A321319 Smallest power of 2 obtainable by partitioning the binary representation of n into consecutive blocks and then summing.

Views

Author

Keywords

Examples

Links

Crossrefs

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.

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Author

Keywords

Comments

Examples

Links

Crossrefs