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Perfect squares (A000290) that can be expressed as the sum of three consecutive triangular numbers (A000217).
4

%I #41 Jul 07 2024 17:32:11

%S 4,64,361,6241,35344,611524,3463321,59923081,339370084,5871850384,

%T 33254804881,575381414521,3258631508224,56381506772644,

%U 319312633001041,5524812282304561,31289379402593764,541375222159074304,3066039868821187801,53049246959306977201,300440617765073810704,5198284826789924691364

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

%C 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 1/2 (3k^2+9k+8) is a perfect square, or equivalently when s>=2 and sqrt(24s^2-15) is congruent to 3 mod 6. Furthermore, with the exception of the first term, the members of this sequence are precisely those perfect squares that are also centered triangular numbers (A005448). For s>=2, the values of s are in A129445, and the corresponding indices of the smallest of the 3 triangular numbers are given in A165517.

%H Vincenzo Librandi, <a href="/A165516/b165516.txt">Table of n, a(n) for n = 1..1000</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_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,98,-98,-1,1).

%F a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).

%F a(n) = 98*a(n-2) - a(n-4) - 30. - _Ant King_, Dec 09 2010

%F a(n) = (1/32)*(10 -3*(sqrt(6)-3) * (5-2*sqrt(6))^n + (2+ sqrt(6)) * (-5-2*sqrt(6))^n -(sqrt(6)-2) *(2*sqrt(6)-5)^n + 3*(3+sqrt(6)) *(5+2*sqrt(6))^n).

%F G.f.: x*(4+60*x-95*x^2+x^4)/((1-x)*(1-10*x+x^2)*(1+10*x+x^2)).

%F 16*a(n) = 5 +9*A072256(n+1) +2*(-1)^n*A054320(n). - _R. J. Mathar_, Apr 28 2020

%e The fourth perfect square that can be expressed as the sum of three consecutive triangular numbers is 6241 (=T63+T64+T65), and hence a(4)=6241.

%t Select[Range[2,1.8 10^7],Mod[Sqrt[24#^2-15],6]==3 &]^2

%t CoefficientList[Series[(4 + 60 x - 95 x^2 + x^4)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Mar 14 2014 *)

%t LinearRecurrence[ {1,98,-98,-1,1}, {4, 64, 361, 6241, 35344}, 50] (* _G. C. Greubel_, Oct 21 2018 *)

%o (PARI) Vec(O(x^66)+x*(4+60*x-95*x^2+x^4)/((1-x)*(1-10*x+x^2)*(1+10*x+x^2))) \\ _Joerg Arndt_, Mar 13 2014

%o (Magma) I:=[4, 64, 361, 6241, 35344]; [n le 5 select I[n] else Self(n-1) + 98*Self(n-2) - 98*Self(n-3) - Self(n-4) + Self(n-5): n in [1..50]]; // _G. C. Greubel_, Oct 21 2018

%Y Cf. A000290, A129445, A005448, A000290, A000217, A165517.

%K easy,nonn

%O 1,1

%A _Ant King_, Sep 25 2009, Oct 01 2009

%E a(1) = 4 added by _N. J. A. Sloane_, Sep 28 2009, at the suggestion of _Alexander R. Povolotsky_

%E a(16)-a(21) added by _Alex Ratushnyak_, Mar 12 2014