A165518 Perfect squares (A000290) that can be expressed as the sum of four consecutive triangular numbers (A000217).

4, 100, 3364, 114244, 3880900, 131836324, 4478554084, 152139002500, 5168247530884, 175568277047524, 5964153172084900, 202605639573839044, 6882627592338442564, 233806732499933208100, 7942546277405390632804, 269812766699283348307204, 9165691521498228451812100, 311363698964240484013304164

Offset: 1

Author

Ant King, Sep 28 2009

Keywords

• easy
• nice
• nonn

Comments

As T(n) + T(n+1) = (n+1)^2 and T(n+2) + T(n+3) = (n+3)^2, it follows that the equation T(n) + T(n+1) + T(n+2) + T(n+3) = s^2 becomes (n+1)^2 + (n+3)^2 = s^2. Hence the solutions to this equation correspond to those Pythagorean triples with shorter legs that differ by two, such as 6^2 + 8^2 = 10^2.

Terms are the squares of the hypotenuses of Pythagorean triangles where other two sides are m and m+2, excepting the initial 4. See A075870. - Richard R. Forberg, Aug 15 2013

Examples

			As the third perfect square that can be expressed as the sum of four consecutive triangular numbers is 3364 = T(39) + T(40) + T(41) + T(42), we have a(3)=3364.
The first term, 4, equals T(-1) + T(0) + T(1) + T(2).
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..600
• 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 (35,-35,1).

Crossrefs

Cf. A000290, A000217, A165516 (squares that can be expressed as the sum of three consecutive triangular numbers), A029549, A075870.

Programs

Magma

I:=[4,100,3364]; [n le 3 select I[n] else 35*Self(n-1) - 35*Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Oct 21 2018

Maple

A165518:=n->(1/2)*(2+(3+2*sqrt(2))^(2*n+1)+(3-2*sqrt(2))^(2*n+1)); seq(A165518(k), k=1..20); # Wesley Ivan Hurt, Oct 24 2013

Mathematica

TriangularNumber[n_]:=1/2 n (n+1); data=Select[Range[10^7],IntegerQ[Sqrt[ TriangularNumber[ # ]+TriangularNumber[ #+1]+TriangularNumber[ #+2]+TriangularNumber[ #+3]]] &];2(#^2+4#+5)&/@data
t={4, 100}; Do[AppendTo[t, 34 t[[-1]] - t[[-2]] - 32], {20}]; t
LinearRecurrence[{35,-35,1},{4,100,3364},20] (* Harvey P. Dale, May 22 2012 *)

PARI

x='x+O('x^50); Vec(4*x*(1-10*x+x^2)/((1-x)*(1-34*x+x^2))) \\ G. C. Greubel, Oct 21 2018

Formula

a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).

a(n) = 34*a(n-1) - a(n-2) - 32.

a(n) = (2 + (3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1))/2.

a(n) = ceiling((1/2)*(2 + (3+2*sqrt(2))^(2n+1))).

G.f.: 4*x*(x^2-10*x+1)/((1-x)*(x^2-34*x+1)).

a(n) = 4*A008844(n-1). - R. J. Mathar, Dec 14 2010

a(n) = A075870(n)^2. - Richard R. Forberg, Aug 15 2013

Extensions

Extended by T. D. Noe, Dec 09 2010