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

Author

M. F. Hasler, Nov 25 2012

Keywords

• nonn

Comments

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

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

From Charles Kusniec, Oct 31 2024: (Start)

a(n) = 1 iff 2n is a square.

a(n) = 2 iff 2n-1 = (2k-1)^2 - 4.

a(n) = 3 iff 2n-1 = (2k)^2 - 9.

Since 2n-1 is odd, 2n-1 = d*D can be written 2n-1 = (x-y)*(x+y) = x^2 - y^2 with integers x and y, from which a(n) is the smallest possible y, which also means smallest possible x and that being x = A377499(n).

Record values occur when 2n-1 is an odd prime (n in A006254), and then a(n) = n-1 (and A377499(n) = n). (End)

From Rémi Guillaume, Mar 12 2025: (Start)

a(n) is half the difference between the "median" divisors of 2n-1.

a(n) and A377499(n) have opposite parities.

a(n) and n have opposite parities. (End)

Examples

			For n = 2, consider divisors of 2n - 1 = 3 which are {1, 3}. The least one greater than or equal to sqrt(3) is 3, the largest one less than or equal to 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 greater than or equal to sqrt(27) is 9, the largest one less than or equal to 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.
		

Links

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

Crossrefs

Cf. A033676, A033677, A056737, A001844, A016754, A377499.

Cf. A006254 (indices of record highs).

Programs

Mathematica

Table[(Divisors[n][[(Length[Divisors[n]] - Boole[IntegerQ[Sqrt[n]]])/2 + 1]] - Divisors[n][[(Length[Divisors[n]] + Boole[IntegerQ[Sqrt[n]]])/2]])/2, {n, 1, 199, 2}] (* Alonso del Arte, Nov 25 2012, corrected March 21 2024, with help from Giorgos Kalogeropoulos *)
A219695[n_] := (d = Divisors[2n - 1]; l = Floor[Length@d/2 + 1]; (d[[l]] - d[[-l]])/2); Array[A219695, 100] (* Giorgos Kalogeropoulos, Mar 15 2024 *)

PARI

A056737(n)={n=divisors(n); n[(2+#n)\2]-n[(1+#n)\2]}
A219695(n)=A056737(2*n-1)/2  \\ M. F. Hasler, Nov 25 2012

Formula

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

a(n) = sqrt(A377499(n)^2 - (2n-1)). - Charles Kusniec, Oct 31 2024