A238008 Numbers n such that n*(n+3)*(n+6) is a triangular number.

-5, -3, 0, 1, 10, 12, 22, 159, 639, 651, 2629

Offset

1,1

Comments

From Joerg Arndt, Feb 27 2014: (Start)

Using x for n, we are looking for integral points on the elliptic curve y*(y+1) == 2 * x*(x+3)*(x+6).

Substituting x --> x/2 and y --> y/2 and dividing the equation by 4 we obtain the Weierstrass form y^2 + 2*y == x^3 + 18*x^2 + 72*x.

Running the Sage program gives the following list of points (x : y : 1):

[(-10 : 8 : 1), (-7 : 5 : 1), (-6 : 0 : 1), (0 : 0 : 1), (2 : 14 : 1), (20 : 128 : 1), (24 : 160 : 1), (29 : 203 : 1), (44 : 350 : 1), (318 : 5830 : 1), (1278 : 46008 : 1), (1302 : 47304 : 1), (5258 : 381920 : 1)].

Dividing all x by 2 gives

[-5, -7/2, -3, 0, 1, 10, 12, 29/2, 22, 159, 639, 651, 2629].

The integral values are the terms of this sequence.

(End)

Links

Table of n, a(n) for n=1..11.

Example

1 is in the sequence because 1*4*7=28 is a triangular number.

Prog

(PARI)
istriang(n)=issquare(8*n+1);
isok(n)=istriang( n*(n+3)*(n+6) );
for (n=-10^6, 10^6, if ( isok(n), print1(n, ", ") ) );
\\ Joerg Arndt, Feb 17 2014
(Sage)
# for the curve y^2 + b1*x*y + b3*y =  x^3 + b2*x^2 + b4*x + b5, use
# EllipticCurve([b1, b2, b3, b4, b5])
# we have       y^2 +           2*y == x^3 + 18*x^2 + 72*x +  0, so need
E=EllipticCurve([0, 18, 2, 72, 0])
E.integral_points()
## Joerg Arndt, Feb 27 2014
(PARI) isok(n) = ispolygonal((n-1)*n*(n+1), 3); \\ Michel Marcus, Mar 05 2014

Crossrefs

Cf. A000217, A165519, A165893.

Sequence in context: A152419 A322758 A077602 * A193547 A144481 A350550

Adjacent sequences: A238005 A238006 A238007 * A238009 A238010 A238011

Keyword

sign,fini,full

Author

Alex Ratushnyak, Feb 16 2014

Extensions

Added the negative terms, Joerg Arndt, Feb 27 2014

Status

approved