A165519 Integers k for which k(k+1)(k+2) is a triangular number.

-2, -1, 0, 1, 4, 5, 9, 56, 636

Offset: 1

Author

Ant King, Sep 28 2009

Keywords

• sign
• fini
• full

Comments

This sequence is complete; there are no other integers k for which k(k+1)(k+2) is a triangular number.

Integers k such that 8*k*(k+1)*(k+2)+1 is a square. - Robert Israel, Nov 07 2014

Examples

			The third triangular number which is a product of three consecutive integers is 4*5*6=120=T(15), but 4 is the fifth integer k for which k(k+1)(k+2) is a triangular number, so a(5)=4.
		

References

• R. K. Guy, "Figurate Numbers", D3 in Unsolved Problems in Number Theory, 2nd ed., New York, Springer-Verlag, 1994, p. 148.
• James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 19.

Links

• S. P. Mohanty, Which triangular numbers are products of three consecutive integers?, Acta Math Hung 58, 31-36 (1991).

Crossrefs

Cf. A000217, A001219.

Programs

Magma

[-2,-1] cat [n: n in [0..1000] | IsSquare(8*n^3+24*n^2 +16*n+1)]; // Vincenzo Librandi, Nov 10 2014

Maple

select(x -> issqr(8*x^3 + 24*x^2 + 16*x+1), [$-2..1000]); # Robert Israel, Nov 07 2014

Mathematica

TriangularNumberQ[k_]:=If[IntegerQ[1/2 (Sqrt[1+8k]-1)],True,False]; Select[Range[750],TriangularNumberQ[ # (#+1)(#+2)] &]
With[{nos=Partition[Range[0,1000],3,1]},Transpose[Select[nos, IntegerQ[ (Sqrt[1+8Times@@#]-1)/2]&]][[1]]] (* Harvey P. Dale, Dec 25 2011 *)

PARI

isok(k) = ispolygonal(k*(k+1)*(k+2), 3); \\ Michel Marcus, Oct 31 2014

Extensions

Initial 0 added by Alexander R. Povolotsky, Sep 29 2009

Initial -2 and -1 added by Alex Ratushnyak, Nov 07 2014