%I #21 Dec 14 2021 10:06:31
%S 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,
%T 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,
%U 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
%N Number of vertices on the axis of symmetry of the symmetric representation of sigma(n) with subparts.
%C The number of middle divisors of n is equal to a(n) - 1.
%C For the definition of "subparts" see A279387.
%H Antti Karttunen, <a href="/A348406/b348406.txt">Table of n, a(n) for n = 1..65537</a>
%F a(n) = 1 + A067742(n).
%e For n = 2, 6 and 10 the symmetric representation of sigma(n) with subparts respectively looks like this:
%e .
%e . _ _ _
%e . _| | | | | |
%e . 2 |_ _| | | | |
%e . _ _| | | |
%e . | _ _| | |
%e . _ _ _| |_| _ _| |
%e . 6 |_ _ _ _| | _ _|
%e . _ _|_|
%e . | _|
%e . _ _ _ _ _| |
%e . 10 |_ _ _ _ _ _|
%e .
%e 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.
%e 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.
%e 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.
%t a[n_] := 1 + DivisorSum[n, 1 &, n/2 <= #^2 < 2*n &]; Array[a, 100] (* _Amiram Eldar_, Oct 17 2021 *)
%o (PARI)
%o A067742(n) = sumdiv(n, d, my(d2 = d^2); n / 2 < d2 && d2 <= n << 1); \\ From A067742
%o A348406(n) = (1 + A067742(n));
%Y Companion of A348364.
%Y Cf. A067742, A071090, A071540, A071562, A071563, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A240542, A279387, A281007, A299761, A303297, A340833, A340847, A346868, A347950, A348327.
%K nonn
%O 1,1
%A _Omar E. Pol_, Oct 17 2021