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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. 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

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

LINKS

Table of n, a(n) for n=1..90.

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: A063956 A128214 A307865 * A267186 A248092 A145105

Adjacent sequences:  A219692 A219693 A219694 * A219696 A219697 A219698

KEYWORD

nonn

AUTHOR

M. F. Hasler, Nov 25 2012

STATUS

approved

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Last modified May 29 20:42 EDT 2020. Contains 334710 sequences. (Running on oeis4.)