A115881 a(n) is the largest positive y satisfying the Diophantine equation x^2=y(y+n). a(n)=0 if there are no solutions.

0, 0, 1, 0, 4, 2, 9, 1, 16, 8, 25, 4, 36, 18, 49, 9, 64, 32, 81, 16, 100, 50, 121, 25, 144, 72, 169, 36, 196, 98, 225, 49, 256, 128, 289, 64, 324, 162, 361, 81, 400, 200, 441, 100, 484, 242, 529, 121, 576, 288, 625, 144, 676, 338, 729, 169, 784, 392, 841, 196

Offset: 1

Author

Giovanni Resta, Feb 02 2006

Keywords

• nonn

Comments

The corresponding least y is given by A067721(n).

Examples

			a(15)=49, since the solutions (x,y) to x^2=y(y+15) are (4,1), (10,5), (18, 12) and (56, 49). The largest y is 49, from (x,y)=(56,49).
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A067721, A115878, A115879, A115880.

Programs

Mathematica

Table[Max[y/.Solve[{x^2==y*(y+n),y>0},{x,y},Integers]],{n,1,100}]/.y->0 (* Vaclav Kotesovec, Jun 26 2014 *)

Python

def A115881(n):
    a, b = divmod(n,4)
    return ((c:=a**2)-(a<<1)+1,(d:=c<<2),c<<1,d+(a<<2)+1)[b] # Chai Wah Wu, Aug 21 2024

Formula

Empirical g.f.: -x^3*(x^9+x^8+2*x^7+4*x^6+x^5+6*x^4+2*x^3+4*x^2+1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, Jun 26 2014

From empirical g.f.: a(n) = 1/2 - n/2 + 11*n^2/64 + (1/4 - 1/32*n^2) * (2*floor(n/4) + 2*floor((n+1)/4) - n + 1) + (1/4 - 5/64*n^2)*(-1)^n. - Vaclav Kotesovec, Jun 26 2014

From Chai Wah Wu, Aug 21 2024: (Start)

a(4*j) = j^2 - 2*j + 1,

a(4*j+1) = 4*j^2,

a(4*j+2) = 2*j^2,

a(4*j+3) = 4*j^2+4*j+1 (see A115880).

(End)