OFFSET
1,1
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.
a(n) are the squares of the hypotenuse for Pythagorean triangles where other two sides are m and m+2, excepting the initial 4. See A075870. - Richard R. Forberg, Aug 15 2013
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 to sequences with linear recurrences with constant coefficients, signature (35,-35,1).
FORMULA
a(n) = 35a(n-1)-35a(n-2)+a(n-3).
a(n) = 34a(n-1)-a(n-2)-32.
a(n) = 1/2 [2+(3+2sqrt(2))^(2n+1)+(3-2sqrt(2))^(2n+1)].
a(n) = ceiling(1/2(2+(3+2sqrt(2))^(2n+1))).
G.f.: 4x(x^2-34x+25)/((1-x)(x^2-34x+1)).
a(n) = 4*A008844(n-1). - R. J. Mathar, Dec 14 2010
a(n) = A075870(n)^2. - Richard R. Forberg, Aug 15 2013
EXAMPLE
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).
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 *)
CROSSREFS
KEYWORD
easy,nice,nonn
AUTHOR
Ant King, Sep 28 2009
EXTENSIONS
Extended by T. D. Noe, Dec 09 2010
STATUS
approved