

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
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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*n1)/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



