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%I #22 Sep 08 2022 08:45:27
%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.
%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
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