A077260 Triangular numbers that are 1/5 of a triangular number.

0, 3, 21, 990, 6786, 318801, 2185095, 102652956, 703593828, 33053933055, 226555027545, 10643263790778, 72950015275686, 3427097886697485, 23489678363743371, 1103514876252799416, 7563603483110089800, 355328363055514714491, 2435456831883085172253

Offset

0,2

Comments

The triangular numbers 5x these are in A077261.

Links

Colin Barker, Table of n, a(n) for n = 0..798

Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.

Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.

Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.

Index entries for linear recurrences with constant coefficients, signature (1,322,-322,-1,1).

Formula

a(n) = b(n)*(b(n)+1)/2 where b(n) = A077259(n).

a(n) = (A000045(A007310(n+1))^2-1)/8. - Vladeta Jovovic, Nov 02 2002. - Definition corrected by R. J. Mathar, Sep 16 2009

G.f.: (-3*x*(x^2+6*x+1))/((x-1)*(x^2-18*x+1)*(x^2+18*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009

a(n) = 322*a(n-2) - a(n-4) + 24. - Vladimir Pletser, Mar 23 2020

Example

Since b(3)=44 -> a(3)=44*45/2=990.

Mathematica

CoefficientList[Series[(-3 x (x^2 + 6 x + 1))/((x - 1) (x^2 - 18 x + 1)*(x^2 + 18 x + 1)), {x, 0, 18}], x] (* Michael De Vlieger, Apr 21 2021 *)
LinearRecurrence[{1, 322, -322, -1, 1}, {0, 3, 21, 990, 6786}, 20] (* Harvey P. Dale, Dec 12 2023 *)

Prog

(PARI) concat(0, Vec(-3*x*(x^2+6*x+1) / ((x-1)*(x^2-18*x+1)*(x^2+18*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

Crossrefs

Cf. A000217, A077259, A077261, A077262.

Sequence in context: A111435 A111438 A058635 * A367999 A290766 A290872

Adjacent sequences: A077257 A077258 A077259 * A077261 A077262 A077263

Keyword

easy,nonn

Author

Bruce Corrigan (scentman(AT)myfamily.com), Nov 01 2002

Status

approved