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A165517
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Indices of the least triangular numbers (A000217) for which three consecutive triangular numbers sum to a perfect square (A000290)
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5
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0, 5, 14, 63, 152, 637, 1518, 6319, 15040, 62565, 148894, 619343, 1473912, 6130877, 14590238, 60689439, 144428480, 600763525, 1429694574, 5946945823, 14152517272, 58868694717, 140095478158, 582740001359, 1386802264320
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OFFSET
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1,2
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COMMENTS
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Those perfect squares that can be expressed as the sum of three consecutive triangular numbers correspond to integer solutions of the equation T(k)+T(k+1)+T(k+2)=s^2, or equivalently to 3k^2+9k+8=2s^2. Hence solutions occur whenever (3k^2+9k+8)/2 is a perfect square, or equivalently when s>=2 and sqrt(24s^2-15) is congruent to 3 mod 6. This sequence returns the index of the smallest of the 3 triangular numbers, the values of s^2 are given in A165516 and, with the exception of the first term, the values of s are in A129445.
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REFERENCES
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Beldon, Tom and Gardiner, Tony; Triangular Numbers and Perfect Squares, The Mathematical Gazette, Vol. 86, No. 507, (November 2002), pp. 423-431.
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LINKS
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Table of n, a(n) for n=1..25.
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FORMULA
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a(n)=a(n-1)+10a(n-2)-10a(n-3)-a(n-4)+a(n-5)
G.f.: x(x^3+x^2-9x-5)/((x-1)(x^4-10x^2+1))
a(n)=10*a(n-2)-a(n-4)+12. [From Zak Seidov, Sep 25 2009]
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EXAMPLE
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The fourth perfect square that can be expressed as the sum of three consecutive triangular numbers is 6241 =T(63)+T(64)+T(65). Hence a(4)=63.
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MATHEMATICA
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TriangularNumber[ n_ ]:=1/2 n (n+1); Select[ Range[ 0, 10^7 ], IntegerQ[ Sqrt[ TriangularNumber[ # ]+TriangularNumber[ #+1 ]+TriangularNumber[ #+2 ] ] ] & ]
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CROSSREFS
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Cf. A000290, A129445, A000217, A165516
Sequence in context: A055488 A177049 A127922 * A197788 A197661 A004030
Adjacent sequences: A165514 A165515 A165516 * A165518 A165519 A165520
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KEYWORD
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easy,nonn
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AUTHOR
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Ant King, Sep 25 2009, Oct 01 2009
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EXTENSIONS
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a(1) = 0 added by N. J. A. Sloane, Sep 28 2009, at the suggestion of Alexander Povolotsky.
More terms from Zak Seidov, Sep 25 2009
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STATUS
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approved
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