A281009 - OEIS

%I #28 Feb 21 2017 21:17:10

%S 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,

%T 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,

%U 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

%N Number of odd divisors of n minus the number of middle divisors of n.

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

%C 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).

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

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

%C For more information of the "subparts" see A279387.

%H Robert Israel, <a href="/A281009/b281009.txt">Table of n, a(n) for n = 1..10000</a>

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

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

%e 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.

%e Other examples (conjectured):

%e 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.

%e 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).

%e 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.

%p N:= 200: # to get a(1)..a(N)

%p A:= Vector(N):

%p for m from 1 to N by 2 do

%p   R:= [seq(k*m,k=1..N/m)];

%p   A[R]:= A[R] + Vector(nops(R),1);

%p od:

%p for m from 1 to N do

%p   R:= [seq(k*m, k= floor(m/2)+1..min(2*m,N/m))];

%p   A[R]:= A[R] - Vector(nops(R),1);

%p od:

%p convert(A,list); # _Robert Israel_, Feb 20 2017

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

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

%K nonn

%O 1,3

%A _Omar E. Pol_, Feb 20 2017