A165517 Indices of the least triangular numbers (A000217) for which three consecutive triangular numbers sum to a perfect square (A000290).

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

Offset: 1

Author

Ant King, Sep 25 2009, Oct 01 2009

Keywords

• easy
• nonn

Comments

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.

Examples

			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.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Tom Beldon and Tony Gardiner, Triangular Numbers and Perfect Squares, The Mathematical Gazette, Vol. 86, No. 507, (2002), pp. 423-431.
• Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Crossrefs

Cf. A000290, A129445, A000217, A165516.

Programs

Magma

I:=[0, 5, 14, 63, 152]; [n le 5 select I[n] else Self(n-1) + 10*Self(n-2) - 10*Self(n-3) - Self(n-4) + Self(n-5): n in [1..50]]; // G. C. Greubel, Oct 21 2018

Mathematica

TriangularNumber[ n_ ]:=1/2 n (n+1);Select[ Range[ 0,10^7 ], IntegerQ[ Sqrt[ TriangularNumber[ # ]+TriangularNumber[ #+1 ]+TriangularNumber[ #+2 ] ] ] & ]
CoefficientList[Series[x*(x^3 + x^2 - 9*x - 5)/((x - 1)*(x^4 - 10*x^2 + 1)), {x,0,50}], x] (* or *) LinearRecurrence[{1,10,-10,-1,1}, {0, 5, 14, 63, 152}, 50] (* G. C. Greubel, Feb 17 2017 *)

PARI

x='x+O('x^50); concat([0], Vec(x*(x^3 + x^2 - 9*x - 5)/((x - 1)*(x^4 - 10*x^2 + 1)))) \\ G. C. Greubel, Feb 17 2017

Formula

a(n) = a(n-1) + 10*a(n-2) - 10*a(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. - Zak Seidov, Sep 25 2009

Extensions

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

More terms from Zak Seidov, Sep 25 2009