A121234 Nonnegative integers k such that k*(k+1)*(k+2)+1 is a square.

0, 2, 4, 55

Offset: 1

Author

Roger Cuculière, Aug 21 2006

Keywords

• fini
• nonn
• full

Comments

The sequence is finite by Thue's theorem n(n + 1)(n + 2) + 1 = n^3 + 3*n^2 + 2*n + 1. The set of k values of integral solutions to the elliptic curve y^2 = n^3 + 3*n^2 + 2*n + 1 (see Magma program) is { -2, -1, 0, 2, 4, 55 }. So the sequence is complete. - Mohamed Bouhamida, Nov 29 2007

Examples

			2 * 3 * 4 + 1 = 25 = 5^2, so 2 is in the sequence.
4 * 5 * 6 + 1 = 121 = 11^2, so 4 is in the sequence.
6 * 7 * 8 + 1 = 337, which is a prime number, so 6 is not in the sequence.
		

Crossrefs

Cf. A258692.

Programs

Magma

P := PolynomialRing(Integers()); {k: k in Sort([ p[1] : p in IntegralPoints(EllipticCurve(n^3 + 3*n^2 + 2*n + 1)) ])}; // Mohamed Bouhamida, Nov 29 2007

Mathematica

Select[Range[0,10^5], IntegerQ[Sqrt[#(#+1)(#+2)+1]]&] (* James C. McMahon, Oct 16 2024 *)

PARI

isok(k) = issquare(k*(k+1)*(k+2)+1);  \\ Altug Alkan, Dec 07 2015