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A348406
Number of vertices on the axis of symmetry of the symmetric representation of sigma(n) with subparts.
3
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
1,1
COMMENTS
The number of middle divisors of n is equal to a(n) - 1.
For the definition of "subparts" see A279387.
LINKS
FORMULA
a(n) = 1 + A067742(n).
EXAMPLE
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.
MATHEMATICA
a[n_] := 1 + DivisorSum[n, 1 &, n/2 <= #^2 < 2*n &]; Array[a, 100] (* Amiram Eldar, Oct 17 2021 *)
PROG
(PARI)
A067742(n) = sumdiv(n, d, my(d2 = d^2); n / 2 < d2 && d2 <= n << 1); \\ From A067742
A348406(n) = (1 + A067742(n));
KEYWORD
nonn
AUTHOR
Omar E. Pol, Oct 17 2021
STATUS
approved