A281009 Number of odd divisors of n minus the number of middle divisors of n.

0, 0, 2, 0, 2, 0, 2, 0, 2, 2, 2, 0, 2, 2, 2, 0, 2, 2, 2, 0, 4, 2, 2, 0, 2, 2, 4, 0, 2, 2, 2, 0, 4, 2, 2, 2, 2, 2, 4, 0, 2, 2, 2, 2, 4, 2, 2, 0, 2, 2, 4, 2, 2, 2, 4, 0, 4, 2, 2, 2, 2, 2, 4, 0, 4, 2, 2, 2, 4, 2, 2, 0, 2, 2, 6, 2, 2, 4, 2, 0, 4, 2, 2, 2, 4, 2, 4, 0, 2, 4, 2, 2, 4, 2, 4, 0, 2, 2, 4, 2, 2, 4, 2, 0, 8

Offset: 1

Author

Omar E. Pol, Feb 20 2017

Keywords

• nonn

Comments

Conjecture 1: a(n) is also twice the number of odd divisors of n greater than sqrt(2*n).

Conjecture 2: a(n) is also the number of odd divisors of n less than sqrt(2*n) that are not middle divisors of n, plus the number of odd divisors of n greater than sqrt(2*n).

Conjecture 3: a(n) is also the total number of equidistant subparts in the symmetric representation of sigma(n).

The "equidistant subparts" are the subparts that are not the "central subparts".

For more information of the "subparts" see A279387.

Examples

			For n = 45 the divisors of 45 are [1, 3, 5, 9, 15, 45]. There are 6 odd divisors, and two of them [5 and 9] are also the middle divisors of 45, so a(45) = 6 - 2 = 4.
Other examples (conjectured):
2) There are two odd divisors of 45 that are greater than the square root of 2*45 = 9.4..., so a(45) = 2*2 = 4.
3) The 45th row of A237593 is [23, 8, 5, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 5, 8, 23], and the 44th row of the same triangle is [23, 8, 4, 3, 2, 1, 1, 2, 2, 1, 1, 2, 3, 4, 8, 23], therefore between both symmetric Dyck paths (described in A237593 and A279387) there are two central subparts [27 and 1] and two pairs of equidistant subparts ([23, 23] and [2, 2]). The total number of equidistant subparts is equal to 4, so a(45) = 4. (the diagram of the symmetric representation of sigma(45) is too large to include).
4) The 45th row of A196020 is [89, 43, 27, 0, 13, 9, 0, 0, 1], hence the 45th row of A280850 is [23, 23, 27, 0, 2, 2, 0, 0, 1]. There are two central subparts [27 and 1] and two pairs of equidistant subparts ([23, 23] and [2, 2]). The total number of equidistant subparts is equal to 4, so a(45) = 4.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001227, A067742, A082647, A131576, A196020, A236104, A237048, A237593, A245092, A249351, A261699, A262626, A279667, A280849, A280850, A280940, A281005, A281007, A281008.

Programs

Maple

N:= 200: # to get a(1)..a(N)
A:= Vector(N):
for m from 1 to N by 2 do
  R:= [seq(k*m,k=1..N/m)];
  A[R]:= A[R] + Vector(nops(R),1);
od:
for m from 1 to N do
  R:= [seq(k*m, k= floor(m/2)+1..min(2*m,N/m))];
  A[R]:= A[R] - Vector(nops(R),1);
od:
convert(A,list); # Robert Israel, Feb 20 2017

Mathematica

Table[Count[#, d_ /; OddQ@ d] - Count[#, d_ /; Sqrt[n/2] <= d < Sqrt[2 n]] &@ Divisors@ n, {n, 120}] (* Michael De Vlieger, Feb 20 2017 *)

Formula

a(n) = A001227(n) - A067742(n).

Conjecture: a(n) = 2*A131576(n).