A251924 Numbers n such that the sum of the triangular numbers T(n) and T(n+1) is equal to a hexagonal number H(m) for some m.

0, 34, 1188, 40390, 1372104, 46611178, 1583407980, 53789260174, 1827251437968, 62072759630770, 2108646576008244, 71631910824649558, 2433376321462076760, 82663163018885960314, 2808114166320660573948, 95393218491883573553950, 3240561314557720840260384

Offset: 1

Author

Colin Barker, Dec 11 2014

Keywords

• nonn
• easy

Comments

Also nonnegative integers x in the solutions to 2*x^2-4*y^2+4*x+2*y+2 = 0, the corresponding values of y being A008844.

First bisection of A076708. [Bruno Berselli, Dec 11 2014]

Examples

			34 is in the sequence because T(34)+T(35) = 595+630 = 1225 = H(25).
		

Links

• Colin Barker, Table of n, a(n) for n = 1..653
• Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Crossrefs

Cf. A000217, A000384, A008844, A076708.

Programs

Mathematica

LinearRecurrence[{35,-35,1},{0,34,1188},20] (* Harvey P. Dale, Feb 04 2019 *)

PARI

concat(0, Vec(2*x^2*(x-17)/((x-1)*(x^2-34*x+1)) + O(x^100)))

Formula

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

G.f.: 2*x^2*(x-17) / ((x-1)*(x^2-34*x+1)).

a(n) = (-8-(4+3*sqrt(2))*(17+12*sqrt(2))^(-n)+(-4+3*sqrt(2))*(17+12*sqrt(2))^n)/8. - Colin Barker, Mar 02 2016