1,1

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

Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), page 427.

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).

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

E.g.f.: exp(x)*(1 + exp(16*x)*(3*cosh(12*sqrt(2)*x) - 2*sqrt(2)*sinh(12*sqrt(2)*x))) - 4. - Stefano Spezia, Nov 08 2025

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).

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

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

(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

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

Sequence in context: A244352 A173987 A052144 * A127776 A266524 A224167

Adjacent sequences: A165515 A165516 A165517 * A165519 A165520 A165521

easy,nice,nonn

Ant King, Sep 28 2009

Extended by T. D. Noe, Dec 09 2010

approved