A115878 a(n) is the number of positive solutions of the Diophantine equation x^2 = y(y+n).

0, 0, 1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 4, 2, 1, 2, 1, 1, 4, 1, 1, 4, 2, 1, 3, 1, 1, 4, 1, 3, 4, 1, 4, 2, 1, 1, 4, 4, 1, 4, 1, 1, 7, 1, 1, 7, 2, 2, 4, 1, 1, 3, 4, 4, 4, 1, 1, 4, 1, 1, 7, 4, 4, 4, 1, 1, 4, 4, 1, 7, 1, 1, 7, 1, 4, 4, 1, 7, 4, 1, 1, 4, 4, 1, 4, 4, 1, 7, 4, 1, 4, 1, 4, 10, 1, 2, 7, 2, 1, 4

Offset: 1

Author

Giovanni Resta, Feb 02 2006

Keywords

• nonn
• easy

Comments

Number of divisors d of n^2 such that d^2 < n^2 and n^2/d == d (mod 4). - Antti Karttunen, Oct 06 2018, based on Robert Israel's Jun 27 2014 comment in A115880.

For odd n, a(n) can be computed from the prime signature. - David A. Corneth, Oct 07 2018

Number of r X s rectangles with integer side lengths such that r + s = n, r < s and (s-r) | (s*r). - Wesley Ivan Hurt, Apr 24 2020

Examples

			a(15) = 4 since there are 4 solutions (x,y) to x^2 = y(y+15), namely (4,1), (10,5), (18, 12) and (56, 49).
Note how each x is obtained from each such divisor pair n2/d and d of n2 as (n2/d - d)/4, when their difference is a positive multiple of four, thus in case of n2 = 15^2 = 225 we get (225/1 - 1)/4 = 56, (225/3 - 3)/4 = 18, (225/5 - 5) = 10 and (225/9 - 9)/4 = 4. - _Antti Karttunen_, Oct 06 2018
a(96) = 10. We compute P, the largest power of 2 dividing n = 96. Then compute min(P, 4) and divide n by it. This gives 96/4 = 24. Then find the number of divisors of 24^2, which is 21. Dividing by 2 rounding down to the nearest integer gives 10, the value of a(96). - _David A. Corneth_, Oct 06 2018
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..18480
• Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000

Crossrefs

Cf. A067721, A115879, A115880, A115881.

Programs

Mathematica

a[n_] := Sum[Boole[d^2 < n^2 && Mod[n^2/d-d, 4] == 0], {d, Divisors[n^2]}];
Array[a, 102] (* Jean-François Alcover, Feb 27 2019, from PARI *)

PARI

A115878(n) = { my(n2 = n^2); sumdiv(n2,d,((d*d)Antti Karttunen, Oct 06 2018

PARI

a(n) = my(v=min(2, valuation(n,2))); numdiv((n>>v)^2)>>1 \\ David A. Corneth, Oct 06 2018

Python

from itertools import takewhile
from sympy import divisors
def A115878(n): return sum(1 for d in takewhile(lambda d:dChai Wah Wu, Aug 21 2024

Formula

From David A. Corneth, Oct 07 2018: (Start)

a((2k+1) * 2^m) = floor(tau((2k + 1) ^ 2) / 2) for m <= 2.

a((2k+1) * 2^m) = (2m - 3) * a(2k+1) + (m-2) for m > 2. (End)

a(n) = Sum_{i=1..floor((n-1)/2)} (1 - ceiling(i*(n-i)/(n-2*i)) + floor(i*(n-i)/(n-2*i))). - Wesley Ivan Hurt, Apr 24 2020