A281005 Numbers n having at least one odd divisor greater than sqrt(2*n).

3, 5, 7, 9, 10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25, 26, 27, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 97, 98, 99, 100, 101, 102, 103, 105

Offset: 1

Author

Omar E. Pol, Feb 06 2017

Keywords

• nonn

Comments

Conjecture 1: also numbers n such that the symmetric representation of sigma(n) has at least one pair of equidistant subparts.

Conjecture 2: the number of pairs of equidistant subparts in the symmetric representation of sigma(k) equals the number of odd divisors of k greater than sqrt(2*k), with k >= 1.

For more information about the subparts see A279387.

Examples

			18 is in the sequence because one of its odd divisors is 9, and 9 is greater than 6, the square root of 2*18.
On the other hand the symmetric representation of sigma(18) has only one part of size 39, which is formed by a central subpart of size 35 and a pair of equidistant subparts [2, 2]. Since there is at least one pair of equidistant subparts, so 18 is in the sequence.
From _Omar E. Pol_, Dec 18 2020: (Start)
The 17th row of triangle A237593 is [9, 4, 2, 1, 1, 1, 1, 2, 4, 9] and the 18th row of the same triangle is [10, 3, 2, 2, 1, 1, 2, 2, 3, 10], so the diagram of the symmetric representation of sigma(18) = 39 is constructed as shown below in figure 1:
.                                     _                                      _
.                                    | |                                    | |
.                                    | |                                    | |
._                                   | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                                    | |                                    | |
.                             _ _ _ _| |                             _ _ _ _| |
.                            |    _ _ _|                            |  _ _ _ _|
.                           _|   |                                 _| | |
.                         _|  _ _|                               _|  _|_|
.                     _ _|  _|                               _ _|  _|    2
.                    |     |  39                            |  _ _|
.                    |  _ _|                                | |_ _|
.                    | |                                    | |    2
.   _ _ _ _ _ _ _ _ _| |                   _ _ _ _ _ _ _ _ _| |
.  |_ _ _ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _ _ _|
.                                                              35
.
.   Figure 1. The symmetric               Figure 2. After the dissection
.   representation of sigma(18)           of the symmetric representation
.   has one part of size 39.              of sigma(18) into layers of
.                                         width 1 we can see three subparts.
.                                         The first layer has one subpart of
.                                         size 35. The second layer has
.                                         two equidistant subparts of size 2,
.                                         so 18 is in the sequence.
(End)
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 1..1000

Crossrefs

Complement of A082662.

Indices of positive terms in A131576.

Cf. A000203, A001227, A067742, A071561, A196020, A235791, A236104, A237048, A237270, A237271, A237571, A237593, A244050, A245092, A249351, A262626, A279387, A280850, A280851, A296508.

Programs

Magma

[k:k in [1..110] | not forall{d:d in Divisors(k)| IsEven(d) or d le Sqrt(2*k)}]; // Marius A. Burtea, Jan 15 2020

Mathematica

Select[Range@ 120, Count[Divisors@ #, d_ /; And[OddQ@ d, d > Sqrt[2 #]]] > 0 &] (* Michael De Vlieger, Feb 07 2017 *)

PARI

isok(n) = my(s=sqrt(2*n)); sumdiv(n, d, (d % 2) && (d > s)) > 0; \\ Michel Marcus, Jan 15 2020