A121234 - OEIS

%I #26 Oct 16 2024 20:33:11

%S 0,2,4,55

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

%C 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

%e 2 * 3 * 4 + 1 = 25 = 5^2, so 2 is in the sequence.

%e 4 * 5 * 6 + 1 = 121 = 11^2, so 4 is in the sequence.

%e 6 * 7 * 8 + 1 = 337, which is a prime number, so 6 is not in the sequence.

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

%o (Magma) P<n> := 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

%o (PARI) isok(k) = issquare(k*(k+1)*(k+2)+1);  \\ _Altug Alkan_, Dec 07 2015

%Y Cf. A258692.

%K fini,nonn,full

%O 1,2

%A _Roger Cuculière_, Aug 21 2006