A348406 Number of vertices on the axis of symmetry of the symmetric representation of sigma(n) with subparts.

2, 2, 1, 2, 1, 3, 1, 2, 2, 1, 1, 3, 1, 1, 3, 2, 1, 2, 1, 3, 1, 1, 1, 3, 2, 1, 1, 3, 1, 3, 1, 2, 1, 1, 3, 2, 1, 1, 1, 3, 1, 3, 1, 1, 3, 1, 1, 3, 2, 2, 1, 1, 1, 3, 1, 3, 1, 1, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 1, 3, 1, 4, 1, 1, 1, 1, 3, 1, 1, 3, 2, 1, 1, 3, 1, 1, 1, 3, 1, 3, 3, 1, 1, 1, 1, 3, 1, 2, 3, 2, 1, 1, 1, 3, 1

Offset: 1

Author

Omar E. Pol, Oct 17 2021

Keywords

• nonn

Comments

The number of middle divisors of n is equal to a(n) - 1.

For the definition of "subparts" see A279387.

Examples

			For n = 2, 6 and 10 the symmetric representation of sigma(n) with subparts respectively looks like this:
.
.           _       _       _
.         _| |     | |     | |
.    2   |_ _|     | |     | |
.               _ _| |     | |
.              |  _ _|     | |
.         _ _ _| |_|    _ _| |
.    6   |_ _ _ _|     |  _ _|
.                   _ _|_|
.                  |  _|
.         _ _ _ _ _| |
.   10   |_ _ _ _ _ _|
.
For n = 2 there are two vertices on the axis of symmetry hence the number of middle divisors of 2 is equal to 2 - 1 = 1.
For n = 6 there are three vertices on the axis of symmetry hence the number of middle divisors of 6 is equal to 3 - 1 = 2.
For n = 10 there is only one vertex on the axis of symmetry hence the number of middle divisors of 10 is equal to 1 - 1 = 0.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Companion of A348364.

Cf. A067742, A071090, A071540, A071562, A071563, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A240542, A279387, A281007, A299761, A303297, A340833, A340847, A346868, A347950, A348327.

Programs

Mathematica

a[n_] := 1 + DivisorSum[n, 1 &, n/2 <= #^2 < 2*n &]; Array[a, 100] (* Amiram Eldar, Oct 17 2021 *)

PARI

A067742(n) = sumdiv(n, d, my(d2 = d^2); n / 2 < d2 && d2 <= n << 1); \\ From A067742
A348406(n) = (1 + A067742(n));

Formula

a(n) = 1 + A067742(n).