Search: a320142 -id:a320142
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A320137
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Numbers that have only one middle divisor.
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+10
8
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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
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OFFSET
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1,2
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COMMENTS
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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.
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)
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LINKS
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EXAMPLE
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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:
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MATHEMATICA
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(* 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]
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]
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CROSSREFS
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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.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A320051
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Square array read by antidiagonals upwards: T(n,k) is the n-th positive integer with exactly k middle divisors, n >= 1, k >= 0.
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+10
3
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3, 5, 1, 7, 2, 6, 10, 4, 12, 72, 11, 8, 15, 144, 120, 13, 9, 20, 288, 180, 1800, 14, 16, 24, 400, 240, 3528, 840, 17, 18, 28, 450, 252, 4050, 1080, 3600, 19, 25, 30, 576, 336, 5184, 1260, 7200, 2520, 21, 32, 35, 648, 360, 7056, 1440, 14112, 5040, 28800, 22, 36, 40, 800, 378, 8100, 1680, 14400, 5544
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OFFSET
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1,1
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COMMENTS
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This is a permutation of the natural numbers.
For the definition of middle divisors see A067742.
Conjecture 1: T(n,k) is also the n-th positive integer j with the property that the difference between the number of partitions of j into an odd number of consecutive parts and the number of partitions of j into an even number of consecutive parts is equal to k.
Conjecture 2: T(n,k) is also the n-th positive integer j with the property that the symmetric representation of sigma(j) has width k on the main diagonal.
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LINKS
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EXAMPLE
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The corner of the square array begins:
3, 1, 6, 72, 120, 1800, 840, 3600, 2520, 28800, ...
5, 2, 12, 144, 180, 3528, 1080, 7200, 5040, ...
7, 4, 15, 288, 240, 4050, 1260, 14112, ...
10, 8, 20, 400, 252, 5184, 1440, ...
11, 9, 24, 450, 336, 7056, ...
13, 16, 28, 576, 360, ...
14, 18, 30, 648, ...
17, 25, 35, ...
19, 32, ...
21, ...
...
In accordance with the conjecture 1, T(1,0) = 3 because there is only one partition of 3 into an odd number of consecutive parts: [3], and there is only one partition of 3 into an even number of consecutive parts: [2, 1], therefore the difference of the number of those partitions is 1 - 1 = 0.
On the other hand, in accordance with the conjecture 2: T(1,0) = 3 because the symmetric representation of sigma(3) = 4 has width 0 on the main diagonal, as shown below:
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In accordance with the conjecture 1, T(1,2) = 6 because there are three partitions of 6 into an odd number of consecutive parts: [6], [3, 2, 1], and there are no partitions of 6 into an even number of consecutive parts, therefore the difference of the number of those partitions is 2 - 0 = 2.
On the other hand, in accordance with the conjecture 2: T(1,2) = 6 because the symmetric representation of sigma(6) = 12 has width 2 on the main diagonal, as shown below:
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CROSSREFS
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The union of the rest of the columns gives A071562.
For more information about the diagrams see A237593.
For tables of partitions into consecutive parts see A286000 and A286001.
Cf. A067742, A240542, A245092, A249351 (widths), A262626, A279286, A280849, A281007, A299761, A299777, A303297, A319529, A319796, A319801, A319802.
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KEYWORD
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AUTHOR
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STATUS
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approved
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A352425
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Irregular triangle read by rows in which row n lists the partitions of n into an odd number of consecutive parts.
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+10
3
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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_|1|
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_|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.
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CROSSREFS
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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.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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A355143
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Product of middle divisors of n, or 0 if there are no middle divisors of n.
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+10
1
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1, 1, 0, 2, 0, 6, 0, 2, 3, 0, 0, 12, 0, 0, 15, 4, 0, 3, 0, 20, 0, 0, 0, 24, 5, 0, 0, 28, 0, 30, 0, 4, 0, 0, 35, 6, 0, 0, 0, 40, 0, 42, 0, 0, 45, 0, 0, 48, 7, 5, 0, 0, 0, 54, 0, 56, 0, 0, 0, 60, 0, 0, 63, 8, 0, 66, 0, 0, 0, 70, 0, 432, 0, 0, 0, 0, 77, 0, 0, 80
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OFFSET
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1,4
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LINKS
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EXAMPLE
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For n = 6 the middle divisors of 6 are 2 and 3, the product of them is 2*3 = 6, so a(6) = 6.
For n = 7 there are no middle divisors of 7, so a(7) = 0.
For n = 8 there is only one middle divisor of 8, the 2, so a(8) = 2.
For n = 72 the middle divisors of 72 are [6, 8, 9], the product of them is 6*8*9 = 432, so a(72) = 432.
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MATHEMATICA
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a[n_] := If[(p = Product[If[Sqrt[n/2] <= d < Sqrt[2*n], d, 1], {d, Divisors[n]}]) == 1 && n > 2, 0, p]; Array[a, 100] (* Amiram Eldar, Jun 21 2022 *)
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PROG
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(PARI) a(n) = my(v=select(x->((x >= sqrt(n/2)) && (x < sqrt(n*2))), divisors(n))); if (#v, vecprod(v), 0); \\ Michel Marcus, Aug 04 2022
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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