

A121234


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


1




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 (bhmd95(AT)yahoo.fr), Nov 29 2007


LINKS

Table of n, a(n) for n=1..4.


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.


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 (bhmd95(AT)yahoo.fr), Nov 29 2007
(PARI) isok(k) = issquare(k*(k+1)*(k+2)+1); \\ Altug Alkan, Dec 07 2015


CROSSREFS

Cf. A258692.
Sequence in context: A018337 A092389 A005274 * A084574 A087621 A062784
Adjacent sequences: A121231 A121232 A121233 * A121235 A121236 A121237


KEYWORD

fini,nonn,full


AUTHOR

Roger CuculiÃ¨re, Aug 21 2006


STATUS

approved



