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Numbers that have only one middle divisor.
8

%I #49 Aug 04 2022 16:10:40

%S 1,2,4,8,9,16,18,25,32,36,49,50,64,81,98,100,121,128,162,169,196,200,

%T 225,242,256,289,324,338,361,392,441,484,512,529,578,625,676,722,729,

%U 784,841,882,961,968,1024,1058,1089,1156,1250,1352,1369,1444,1458,1521,1681,1682,1849,1922,1936,2025,2048,2116

%N Numbers that have only one middle divisor.

%C 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.

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

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

%C From _Hartmut F. W. Hoft_, May 24 2022: (Start)

%C 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.

%C 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).

%C 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.

%C 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.

%C 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.

%C 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:

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

%C 9 18 36

%C 25 50 100 200

%C 49 98 196 392 784

%C 81 162 324

%C 121 242 484 968 1936 3872

%C 169 338 676 1352 2704 5408 10816

%C 225

%C 289 578 1156 2312 4624 9248 18496 36992

%C 361 722 1444 2888 5776 11552 23104 46208

%C 441 882

%C 529 1058 2116 4232 8464 16928 33856 67712 135424

%C 625 1250 2500 5000

%C 729 1458 2916

%C 841 1682 3364 6728 13456 26912 53824 107648 215296

%C ...

%C (End)

%H Hartmut F. W. Hoft, <a href="/A320137/a320137.pdf">Proofs of conjectures 2 and 3</a>

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

%e 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.

%e 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:

%e .

%e . _ _ _ _ _ 5

%e . |_ _ _ _ _|

%e . |_ _ 3

%e . |_ |

%e . |_|_ _ 5

%e . | |

%e . | |

%e . | |

%e . | |

%e . |_|

%e .

%t (* computation based on counts of divisors *)

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

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

%t a320137D[2116]

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

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

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

%t a320137W[2116]

%t (* _Hartmut F. W. Hoft_, May 24 2022 *)

%Y Column 1 of A320051.

%Y First differs from A028982 at a(14).

%Y For the definition of middle divisors see A067742.

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

%Y Cf. A016754, A028982, A249223, A259417.

%K nonn

%O 1,2

%A _Omar E. Pol_, Oct 06 2018