1,2

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.

Beldon, Tom and Gardiner, Tony; Triangular Numbers and Perfect Squares, The Mathematical Gazette, Vol. 86, No. 507, (November 2002), pp. 423-431.

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 (zakseidov(AT)yahoo.com), Sep 25 2009]

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.

TriangularNumber[ n_ ]:=1/2 n (n+1); Select[ Range[ 0, 10^7 ], IntegerQ[ Sqrt[ TriangularNumber[ # ]+TriangularNumber[ #+1 ]+TriangularNumber[ #+2 ] ] ] & ]

Cf. A000290, A129445, A000217, A165516

Sequence in context: A073541 A055488 A127922 this_sequence A004030 A166795 A128102

Adjacent sequences: A165514 A165515 A165516 this_sequence A165518 A165519 A165520

easy,nonn

Ant King (mathstutoring(AT)ntlworld.com), Sep 25 2009, Oct 01 2009

a(1) = 0 added by N. J. A. Sloane, Sep 28 2009, at the suggestion of Alexander Povolotsky.

More terms from Zak Seidov (zakseidov(AT)yahoo.com), Sep 25 2009