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A121234
Nonnegative integers k such that k*(k+1)*(k+2)+1 is a square.
2
OFFSET
1,2
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
EXAMPLE
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.
MATHEMATICA
Select[Range[0, 10^5], IntegerQ[Sqrt[#(#+1)(#+2)+1]]&] (* James C. McMahon, Oct 16 2024 *)
PROG
(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
(PARI) isok(k) = issquare(k*(k+1)*(k+2)+1); \\ Altug Alkan, Dec 07 2015
CROSSREFS
Cf. A258692.
Sequence in context: A092389 A341733 A005274 * A084574 A358239 A087621
KEYWORD
fini,nonn,full
AUTHOR
Roger Cuculière, Aug 21 2006
STATUS
approved