A219695 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.

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, 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, 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

Offset

1,3

Comments

We consider 2n - 1 which has only odd divisors, so any difference among them is always even.

Links

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Formula

a(n) = A056737(2n - 1)/2 = (A033677(2n - 1) - A033676(2n - 1))/2.

a(n) = 0 if and only if 2n - 1 is a square.

Example

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.
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.
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.

Mathematica

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 *)

Prog

(PARI) A219695(n)=A056737(2*n-1)/2  \\ M. F. Hasler, Nov 25 2012

Crossrefs

Cf. A033676, A033677.

Sequence in context: A351078 A128214 A307865 * A364575 A267186 A248092

Adjacent sequences: A219692 A219693 A219694 * A219696 A219697 A219698

Keyword

nonn

Author

M. F. Hasler, Nov 25 2012

Status

approved