A321318 -id:A321318 - cp's OEIS frontend

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

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.

A331851 a(n) is the number of distinct values obtained by partitioning the binary representation of n into consecutive blocks, and then multiplying the numbers represented by the blocks.

Original entry on oeis.org

1, 1, 2, 2, 2, 4, 3, 3, 2, 5, 4, 7, 3, 7, 4, 5, 2, 6, 5, 9, 4, 7, 7, 11, 3, 9, 7, 11, 4, 11, 6, 7, 2, 7, 6, 11, 5, 11, 9, 14, 4, 11, 7, 15, 7, 15, 11, 17, 3, 11, 9, 13, 7, 15, 11, 19, 4, 14, 11, 19, 6, 17, 8, 11, 2, 8, 7, 13, 6, 13, 11, 17, 5, 10, 11, 21, 9

Offset: 0

Rémy Sigrist, Jan 29 2020

This sequence is a variant of A321318.

			For n = 6:
- the binary representation of 6 is "110",
- we can split it in 4 ways:
      "110" -> 6
      "1" and "10" -> 1*2 = 2
      "11" and "0" -> 3*0 = 0
      "1" and "1" and "0" -> 1*1*0 = 0
- we have 3 distinct values,
- hence a(6) = 3.

Rémy Sigrist, PARI program for A331851
Index entries for sequences related to binary expansion of n

Cf. A321318 (additive variant).
Cf. A331852 (XOR variant), A331853 (AND variant), A331854 (OR variant).
Cf. A331855 (reverse variant).

PARI
```
See Links section.
```

a(2^k) = 2 for any k > 0.

a(2^k+1) = k+2 for any k > 1.

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.

A321320 Largest power of 2 obtainable by partitioning the binary representation of n into consecutive blocks and then summing.

Original entry on oeis.org

1, 2, 2, 4, 2, 2, 4, 8, 2, 4, 4, 2, 4, 4, 8, 16, 2, 4, 4, 4, 4, 8, 8, 2, 4, 8, 8, 4, 8, 8, 16, 32, 2, 4, 4, 8, 4, 8, 8, 4, 4, 4, 8, 8, 8, 16, 16, 2, 4, 8, 8, 8, 8, 8, 16, 4, 8, 16, 16, 8, 16, 16, 32, 64, 2, 4, 4, 8, 4, 8, 8, 8, 4, 8, 8, 16, 8, 16, 16, 4, 4, 8

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 largest power of 2 is 4.

E. Berlekamp, J. Buhler, Puzzle 6, Puzzles column, Emissary Fall (2011) 9.
Steve Butler, Ron Graham, and Richard Strong, Inserting plus signs and adding, Amer. Math. Monthly 123 (3) (2016), 274-279.
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).

Cf. A321318, A321319, A321321.

A306921 Number of ways of breaking the binary expansion of n into consecutive blocks with no leading zeros.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 5, 5, 8, 8, 9, 9, 12, 12, 8, 8, 12, 12, 12, 12, 16, 16, 6, 6, 10, 10, 12, 12, 16, 16, 12, 12, 18, 18, 18, 18, 24, 24, 10, 10, 16, 16, 18, 18, 24, 24, 16, 16, 24, 24, 24, 24, 32, 32, 7, 7, 12, 12, 15, 15, 20, 20, 16

Offset: 0

Peter Kagey, Mar 16 2019

The number 0 is not considered to have a leading zero.
a(n) >= 2^(A000120(n) - 1).
a(2^n - 1) = 2^(n-1) for n > 0.
a(2^n) = n+1.
Conjecture: n appears A067824(n) times for n > 1.

			For n = 13 the a(13) = 6 partitions are [1101], [1, 101], [110, 1], [1, 10, 1], [11, 0, 1], and [1, 1, 0, 1].
Notice that [1, 1, 01] and [11, 01] are not valid partitions because the last part has a leading zero.

Peter Kagey, Table of n, a(n) for n = 0..8192

Cf. A000120, A067824, A074206, A066099.

A321318 gives number of distinct sums of such partitions.

From Charlie Neder, May 08 2019: (Start)
If n = k*2^e + {0,1} with k odd and e > 0, then a(n) = a(k)*(e+1).
Proof: Each partition of n is uniquely determined by a partition of k (call it K) and a choice of some number, from 0 to e, of trailing digits to append to the final part in K, since any remaining digits must appear as singletons. The conjecture follows, since each ordered factorization of a number m produces two numbers n such that a(n) = m, one of each parity, and A067824(n) = 2*A074206(n).
Corollary: For n >= 1, a(2n) = a(2n+1) = Product{k+1 | k in row n of A066099}. (End)

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.

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A331851 a(n) is the number of distinct values obtained by partitioning the binary representation of n into consecutive blocks, and then multiplying the numbers represented by the blocks.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

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

A321320 Largest power of 2 obtainable by partitioning the binary representation of n into consecutive blocks and then summing.

Views

Author

Keywords

Examples

Links

Crossrefs

A306921 Number of ways of breaking the binary expansion of n into consecutive blocks with no leading zeros.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs