%I
%S 0,1,2,3,0,5,6,1,8,9,2,11,0,3,14,15,4,1,18,5,20,21,2,23,0,7,26,3,8,29,
%T 30,1,4,33,10,35,36,5,2,39,0,41,6,13,44,3,14,7,48,1,50,51,4,53,54,17,
%U 56,9,2,5,0,19,10,63,20,65,6,3,68,69,22,1,12,7,74,75,4,13,78,25,8,81,2,83,0,5,86,9,28,89
%N For odd numbers 2n  1, half the difference between the largest divisor not exceeding the square root, and the least divisor not less than the square root.
%C We consider 2n  1 which has only odd divisors, so any difference among them is always even.
%F a(n) = A056737(2n  1)/2 = (A033677(2n  1)  A033676(2n  1))/2.
%F a(n) = 0 if and only if 2n  1 is a square.
%e For n = 2, consider divisors of 2n  1 = 3 which are {1, 3}. The least one >= sqrt(3) is 3, the largest one <= sqrt(3) is 1; whence a(2) = (3  1)/2 = 1.
%e For n = 14, consider divisors of 2n  1 = 27 which are {1, 3, 9, 27}. The least one >= sqrt(27) is 9, the largest one <= sqrt(27) is 3; whence a(14) = (9  3)/2 = 3.
%e For n = 1, 5, 13, 25,..., the number 2n  1 equals the square 1, 9, 25, 49,...; so the two beforementioned "median divisors" coincide with the square root, and a(n) = 0/2 = 0.
%t Table[(Divisors[n][[Length[Divisors[n]]/2 + 1]]  Divisors[n][[Length[Divisors[n]]/2]])/2, {n, 1, 99, 2}] (* _Alonso del Arte_, Nov 25 2012 *)
%o (PARI) A219695(n)=A056737(2*n1)/2 \\  _M. F. Hasler_, Nov 25 2012
%Y Cf. A033676, A033677.
%K nonn
%O 1,3
%A _M. F. Hasler_, Nov 25 2012
