OFFSET
1,1
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
REFERENCES
Guy, R. K.; "Figurate Numbers", D3 in Unsolved Problems in Number Theory, 2nd ed., New York, Springer-Verlag, 1994, p. 148.
EXAMPLE
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.
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 *)
PROG
(PARI) isok(k) = ispolygonal(k*(k+1)*(k+2), 3); \\ Michel Marcus, Oct 31 2014
(Magma) [-2, -1] cat [n: n in [0..1000] | IsSquare(8*n^3+24*n^2 +16*n+1)]; // Vincenzo Librandi, Nov 10 2014
CROSSREFS
KEYWORD
sign,fini,full
AUTHOR
Ant King, Sep 28 2009
EXTENSIONS
Initial 0 added by Alexander R. Povolotsky, Sep 29 2009
Initial -2 and -1 added by Alex Ratushnyak, Nov 07 2014
STATUS
approved