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%I #27 Sep 08 2022 08:45:47
%S -2,-1,0,1,4,5,9,56,636
%N Integers k for which k(k+1)(k+2) is a triangular number.
%C This sequence is complete; there are no other integers k for which k(k+1)(k+2) is a triangular number.
%C Integers k such that 8*k*(k+1)*(k+2)+1 is a square. - _Robert Israel_, Nov 07 2014
%D Guy, R. K.; "Figurate Numbers", D3 in Unsolved Problems in Number Theory, 2nd ed., New York, Springer-Verlag, 1994, p. 148.
%e The third triangular number which is a product of three consecutive integers is 4*5*6=120=T(15), but 4 is the fifth integer k for which k(k+1)(k+2) is a triangular number, so a(5)=4.
%p select(x -> issqr(8*x^3 + 24*x^2 + 16*x+1), [$-2..1000]); # _Robert Israel_, Nov 07 2014
%t TriangularNumberQ[k_]:=If[IntegerQ[1/2 (Sqrt[1+8k]-1)],True,False]; Select[Range[750],TriangularNumberQ[ # (#+1)(#+2)] &]
%t With[{nos=Partition[Range[0,1000],3,1]},Transpose[Select[nos, IntegerQ[ (Sqrt[1+8Times@@#]-1)/2]&]][[1]]] (* _Harvey P. Dale_, Dec 25 2011 *)
%o (PARI) isok(k) = ispolygonal(k*(k+1)*(k+2), 3); \\ _Michel Marcus_, Oct 31 2014
%o (Magma) [-2,-1] cat [n: n in [0..1000] | IsSquare(8*n^3+24*n^2 +16*n+1)]; // _Vincenzo Librandi_, Nov 10 2014
%Y Cf. A000217, A001219.
%K sign,fini,full
%O 1,1
%A _Ant King_, Sep 28 2009
%E Initial 0 added by _Alexander R. Povolotsky_, Sep 29 2009
%E Initial -2 and -1 added by _Alex Ratushnyak_, Nov 07 2014
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