login
A115878
a(n) is the number of positive solutions of the Diophantine equation x^2 = y(y+n).
10
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
1,9
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
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
EXAMPLE
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
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 *)
PROG
(PARI) A115878(n) = { my(n2 = n^2); sumdiv(n2, d, ((d*d)<n2)&&(0==(((n2/d)-d)%4))); }; \\ 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:d<n, divisors(n**2)) if not (d-n**2//d)&3) # Chai Wah Wu, Aug 21 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Giovanni Resta, Feb 02 2006
STATUS
approved