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A121234 Nonnegative integers k such that k*(k+1)*(k+2)+1 is a square. 1
0, 2, 4, 55 (list; graph; refs; listen; history; text; internal format)
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 Cuculière, Aug 21 2006

STATUS

approved

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Last modified June 19 19:03 EDT 2019. Contains 324222 sequences. (Running on oeis4.)