A165518 - OEIS

%I #40 Feb 16 2022 23:36:13

%S 4,100,3364,114244,3880900,131836324,4478554084,152139002500,

%T 5168247530884,175568277047524,5964153172084900,202605639573839044,

%U 6882627592338442564,233806732499933208100,7942546277405390632804,269812766699283348307204,9165691521498228451812100,311363698964240484013304164

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

%C 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.

%C 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

%H Harvey P. Dale, <a href="/A165518/b165518.txt">Table of n, a(n) for n = 1..600</a>

%H Tom Beldon and Tony Gardiner, <a href="http://www.jstor.org/stable/3621134">Triangular Numbers and Perfect Squares</a>, The Mathematical Gazette, Vol. 86, No. 507, (2002), pp. 423-431.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).

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

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

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

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

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

%F a(n) = 4*A008844(n-1). - _R. J. Mathar_, Dec 14 2010

%F a(n) = A075870(n)^2.  - _Richard R. Forberg_, Aug 15 2013

%e 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.

%e The first term, 4, equals T(-1) + T(0) + T(1) + T(2).

%p 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

%t 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 t={4, 100}; Do[AppendTo[t, 34 t[[-1]] - t[[-2]] - 32], {20}]; t

%t LinearRecurrence[{35,-35,1},{4,100,3364},20] (* _Harvey P. Dale_, May 22 2012 *)

%o (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

%o (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

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

%K easy,nice,nonn

%O 1,1

%A _Ant King_, Sep 28 2009

%E Extended by _T. D. Noe_, Dec 09 2010