A320051 -id:A320051

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A320137

Numbers that have only one middle divisor.

+10
8

1, 2, 4, 8, 9, 16, 18, 25, 32, 36, 49, 50, 64, 81, 98, 100, 121, 128, 162, 169, 196, 200, 225, 242, 256, 289, 324, 338, 361, 392, 441, 484, 512, 529, 578, 625, 676, 722, 729, 784, 841, 882, 961, 968, 1024, 1058, 1089, 1156, 1250, 1352, 1369, 1444, 1458, 1521, 1681, 1682, 1849, 1922, 1936, 2025, 2048, 2116

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

Conjecture 1: sequence consists of numbers k with the property that the difference between the number of partitions of k into an odd number of consecutive parts and the number of partitions of k into an even number of consecutive parts is equal to 1.

Conjecture 2: sequence consists of numbers k with the property that the symmetric representation of sigma(k) has width 1 on the main diagonal.

Conjecture 3: all powers of 2 are in the sequence.

From Hartmut F. W. Hoft, May 24 2022: (Start)

Every number in this sequence is a square or twice a square, i.e., this sequence is a subsequence of A028982, and conjectures 2 and 3 are true (see the link for proofs). Furthermore, all odd numbers in this sequence are squares and form subsequences of A016754 and of A319529.

Every number k in this sequence has the form k = 2^m * q^2, m >= 0, q >= 1 odd, where for any divisor e of q^2 smaller than the largest divisor of q^2 that is less than or equal to row(q^2) = floor((sqrt(8*q^2 + 1) - 1)/2) the inequalities 2^(m+1) * e < row(n) hold (see the link for a proof).

The smallest odd square not in this sequence is 1225 = 35^2 = (5*7)^2 since it has the 3 middle divisors 25, 35, 49 and the width of the symmetric representation of sigma(1225) at the diagonal equals 3. However, the squares of odd primes in this sequence are a subsequence of A259417.

The smallest even square not in this sequence is 144 = 12^2 = (2*2*3)^2 since it has the 3 middle divisors 9, 12, 16 and the width of the symmetric representation of sigma(144) at the diagonal equals 3.

The smallest twice square not in this sequence is 72 = 2 * (2*3)^2 = 2^3 * 3^2 since it has the 3 middle divisors 6, 8, 9 and the width of the symmetric representation of sigma(72) at the diagonal equals 3.

Apart from the powers of 2 in the infinite first row, the numbers in the sequence can be arranged as an irregular triangle with each row containing the finitely many numbers q^2, 2 * q^2, 4 * q^2, ..., 2^m * q^2 satisfying the condition stated above, as shown below:

1 2 4 8 16 32 64 128 256 ...

9 18 36

25 50 100 200

49 98 196 392 784

81 162 324

121 242 484 968 1936 3872

169 338 676 1352 2704 5408 10816

225

289 578 1156 2312 4624 9248 18496 36992

361 722 1444 2888 5776 11552 23104 46208

441 882

529 1058 2116 4232 8464 16928 33856 67712 135424

625 1250 2500 5000

729 1458 2916

841 1682 3364 6728 13456 26912 53824 107648 215296

...

(End)

LINKS

Table of n, a(n) for n=1..62.

Hartmut F. W. Hoft, Proofs of conjectures 2 and 3

EXAMPLE

9 is in the sequence because 9 has only one middle divisor: 3.

On the other hand, in accordance with the first conjecture, 9 is in the sequence because there are two partitions of 9 into an odd number of consecutive parts: [9], [4, 3, 2], and there is only one partition of 9 into an even number of consecutive parts: [5, 4], therefore the difference of the number of those partitions is 2 - 1 = 1.

On the other hand, in accordance with the second conjecture, 9 is in the sequence because the symmetric representation of sigma(9) = 13 has width 1 on the main diagonal, as shown below in the first quadrant:

. _ _ _ _ _ 5

. |_ _ _ _ _|

. |_ _ 3

. |_ |

. |_|_ _ 5

. | |

. |_|

MATHEMATICA

(* computation based on counts of divisors *)

middleDiv[n_] := Select[Divisors[n], Sqrt[n/2]<=#<Sqrt[2n]&]

a320137D[n_] := Select[Range[n], Length[middleDiv[#]]==1&]

a320137D[2116]

(* computation based on A237048 and A249223 for width at diagonal *)

a249223[n_] := Drop[FoldList[Plus, 0, Map[(-1)^(#+1) a237048[n, #]&, Range[Floor[(Sqrt[8n+1]-1)/2]]]], 1]

a320137W[n_] := Select[Range[n], Last[a249223[#]]==1&]

a320137W[2116]

(* Hartmut F. W. Hoft, May 24 2022 *)

CROSSREFS

Column 1 of A320051.

First differs from A028982 at a(14).

For the definition of middle divisors see A067742.

Cf. A000079, A071561, A071562, A237048, A237593, A240542, A245092, A249351 (widths), A279286, A279387, A280849, A281007, A299761, A299777, A303297, A319529, A319796, A319801, A319802, A320142.

Cf. A016754, A028982, A249223, A259417.

KEYWORD

nonn

AUTHOR

Omar E. Pol, Oct 06 2018

STATUS

approved

A320142

Numbers that have exactly two middle divisors.

+10
5

6, 12, 15, 20, 24, 28, 30, 35, 40, 42, 45, 48, 54, 56, 60, 63, 66, 70, 77, 80, 84, 88, 90, 91, 96, 99, 104, 108, 110, 112, 117, 126, 130, 132, 135, 140, 143, 150, 153, 154, 156, 160, 165, 168, 170, 176, 182, 187, 190, 192, 195, 198, 204, 208, 209, 210, 216, 220, 221, 224, 228, 231, 234, 238, 247, 255, 260

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Conjecture 1: numbers k with the property that the difference between the number of partitions of k into an odd number of consecutive parts and the number of partitions of k into an even number of consecutive parts is equal to 2.

Conjecture 2: numbers k with the property that symmetric representation of sigma(k) has width 2 on the main diagonal.

By the theorem in A067742 conjecture 2 is true. - Hartmut F. W. Hoft, Aug 18 2024

LINKS

Table of n, a(n) for n=1..67.

EXAMPLE

15 is in the sequence because 15 has two middle divisors: 3 and 5.

On the other hand, in accordance with the first conjecture, 15 is in the sequence because there are three partitions of 15 into an odd number of consecutive parts: [15], [8, 7], [5, 4, 3, 2, 1], and there is only one partition of 15 into an even number of consecutive parts: [8, 7], therefore the difference of the number of those partitions is 3 - 1 = 2.

On the other hand, in accordance with the second conjecture, 15 is in the sequence because the symmetric representation of sigma(15) = 24 has width 2 on the main diagonal, as shown below in the fourth quadrant:

. _

. | |

. _ _ _|_|

. _ _| | 8

. | _|

. _| _|

. |_ _| 8

. |

. _ _ _ _ _ _ _ _|

. |_ _ _ _ _ _ _ _|

. 8

MATHEMATICA

a320142Q[k_] := Length[Select[Divisors[k], k/2<=#^2<2k&]]==2

a320142[n_] := Select[Range[n], a320142Q]

a320142[260] (* Hartmut F. W. Hoft, Aug 20 2024 *)

CROSSREFS

Column 2 of A320051.

First differs from A001284 at a(19).

For the definition of middle divisors see A067742.

Cf. A071561, A071562, A237048, A237593, A240542, A245092, A249351 (widths), A279286, A279387, A280849, A281007, A299761, A299777, A303297, A319529, A319796, A319801, A319802, A320137.

KEYWORD

nonn

AUTHOR

Omar E. Pol, Oct 06 2018

STATUS

approved

A352425

Irregular triangle read by rows in which row n lists the partitions of n into an odd number of consecutive parts.

+10
3

1, 2, 3, 4, 5, 6, 3, 2, 1, 7, 8, 9, 4, 3, 2, 10, 11, 12, 5, 4, 3, 13, 14, 15, 6, 5, 4, 5, 4, 3, 2, 1, 16, 17, 18, 7, 6, 5, 19, 20, 6, 5, 4, 3, 2, 21, 8, 7, 6, 22, 23, 24, 9, 8, 7, 25, 7, 6, 5, 4, 3, 26, 27, 10, 9, 8, 28, 7, 6, 5, 4, 3, 2, 1, 29, 30, 11, 10, 9, 8, 7, 6, 5, 4

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

Conjecture: the total number of parts in all partitions of n into an odd number of consecutive parts equals the sum of odd divisors of n that are <= A003056(n). In other words: row n has A341309(n) terms.

The first partition with 2*m - 1 parts appears in the row A000384(m), m >= 1.

LINKS

Table of n, a(n) for n=1..84.

EXAMPLE

Triangle begins:

[1];

[2];

[3],

[4];

[5];

[6], [3, 2, 1];

[7];

[8];

[9], [4, 3, 2];

[10];

[11];

[12], [5, 4, 3];

[13];

[14];

[15], [6, 5, 4], [5, 4, 3, 2, 1];

[16];

[17];

[18], [7, 6, 5];

[19];

[20], [6, 5, 4, 3, 2];

[21], [8, 7, 6];

[22];

[23];

[24], [9, 8, 7];

[25], [7, 6, 5, 4, 3];

[26];

[27], [10, 9, 8];

[28], [7, 6, 5, 4, 3, 2, 1];

...

In the diagram below the m-th staircase walk starts at row A000384(m).

The number of horizontal line segments in the n-th row equals A082647(n), the number of partitions of n into an odd number of consecutive parts, so we can find such partitions as follows: consider the vertical blocks of numbers that start exactly in the n-th row of the diagram, for example: for n = 15 consider the vertical blocks of numbers that start exactly in the 15th row. They are [15], [6, 5, 4]. [5, 4, 3, 2, 1], equaling the 15th row of the above triangle.

_|1|

_|2 |

_|3 |

_|4 |

_|5 _|

_|6 |3|

_|7 |2|

_|8 _|1|

_|9 |4 |

_|10 |3 |

_|11 _|2 |

_|12 |5 |

_|13 |4 |

_|14 _|3 _|

_|15 |6 |5|

_|16 |5 |4|

_|17 _|4 |3|

_|18 |7 |2|

_|19 |6 _|1|

_|20 _|5 |6 |

_|21 |8 |5 |

_|22 |7 |4 |

_|23 _|6 |3 |

_|24 |9 _|2 |

_|25 |8 |7 |

_|26 _|7 |6 |

_|27 |10 |5 _|

|28 |9 |4 |7|

...

The diagram is infinite.

For more information about the diagram see A286000.

CROSSREFS

Subsequence of A299765.

Row sums give A352257.

Column 1 gives A000027.

Records give A000027.

Row n contains A082647(n) of the mentioned partitions.

Cf. A000384, A003056, A067742, A204217, A237048, A237591, A237593, A240542, A245092, A285574, A285901, A286000, A286001, A320051, A320137, A320142, A341309, A351824.

KEYWORD

nonn,tabf

AUTHOR

Omar E. Pol, Mar 15 2022

STATUS

approved

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