login
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A165516 Perfect squares (A000290) that can be expressed as the sum of three consecutive triangular numbers (A000217). 4
4, 64, 361, 6241, 35344, 611524, 3463321, 59923081, 339370084, 5871850384, 33254804881, 575381414521, 3258631508224, 56381506772644, 319312633001041, 5524812282304561, 31289379402593764, 541375222159074304, 3066039868821187801, 53049246959306977201, 300440617765073810704, 5198284826789924691364 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

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 (1,98,-98,-1,1).

FORMULA

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

a(n) = 98*a(n-2) - a(n-4) - 30. - Ant King, Dec 09 2010

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).

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

16*a(n) = 5 +9*A072256(n+1) +2*(-1)^n*A054320(n). - R. J. Mathar, Apr 28 2020

EXAMPLE

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.

MATHEMATICA

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

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 *)

LinearRecurrence[{1, 98, -98, -1, 1}, {4, 64, 361, 6241, 35344}, 50] (* G. C. Greubel, Oct 21 2018 *)

PROG

(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

(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

CROSSREFS

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

Sequence in context: A264055 A222557 A249483 * A064935 A030098 A087045

Adjacent sequences:  A165513 A165514 A165515 * A165517 A165518 A165519

KEYWORD

easy,nonn

AUTHOR

Ant King, Sep 25 2009, Oct 01 2009

EXTENSIONS

a(1) = 4 added by N. J. A. Sloane, Sep 28 2009, at the suggestion of Alexander R. Povolotsky

a(16)-a(21) added by Alex Ratushnyak, Mar 12 2014

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified May 30 18:38 EDT 2020. Contains 334728 sequences. (Running on oeis4.)