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A165519
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Integers k for which k(k+1)(k+2) is a triangular number.
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1
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OFFSET
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1,1
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COMMENTS
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This sequence is complete; there are no other integers k for which k(k+1)(k+2) is a triangular number.
Integers k such that 8*k*(k+1)*(k+2)+1 is a square. - Robert Israel, Nov 07 2014
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REFERENCES
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Guy, R. K.; "Figurate Numbers", D3 in Unsolved Problems in Number Theory, 2nd ed., New York, Springer-Verlag, 1994, p. 148.
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LINKS
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EXAMPLE
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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.
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MAPLE
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select(x -> issqr(8*x^3 + 24*x^2 + 16*x+1), [$-2..1000]); # Robert Israel, Nov 07 2014
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MATHEMATICA
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TriangularNumberQ[k_]:=If[IntegerQ[1/2 (Sqrt[1+8k]-1)], True, False]; Select[Range[750], TriangularNumberQ[ # (#+1)(#+2)] &]
With[{nos=Partition[Range[0, 1000], 3, 1]}, Transpose[Select[nos, IntegerQ[ (Sqrt[1+8Times@@#]-1)/2]&]][[1]]] (* Harvey P. Dale, Dec 25 2011 *)
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PROG
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(PARI) isok(k) = ispolygonal(k*(k+1)*(k+2), 3); \\ Michel Marcus, Oct 31 2014
(Magma) [-2, -1] cat [n: n in [0..1000] | IsSquare(8*n^3+24*n^2 +16*n+1)]; // Vincenzo Librandi, Nov 10 2014
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CROSSREFS
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KEYWORD
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sign,fini,full
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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