A251963 Numbers n such that the sum of the triangular numbers T(n) and T(n+1) is equal to an octagonal number N(m) for some m.

0, 14, 208, 2910, 40544, 564718, 7865520, 109552574, 1525870528, 21252634830, 296011017104, 4122901604638, 57424611447840, 799821658665134, 11140078609864048, 155161278879431550, 2161117825702177664, 30100488280951055758, 419245718107612602960

Offset: 1

Author

Colin Barker, Dec 11 2014

Keywords

• nonn
• easy

Comments

Also nonnegative integers x in the solutions to 2*x^2-6*y^2+4*x+4*y+2+2 = 0, the corresponding values of y being A046184.

Examples

			14 is in the sequence because T(14)+T(15) = 105+120 = 225 = N(9).
		

Links

• Colin Barker, Table of n, a(n) for n = 1..874
• Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Crossrefs

Cf. A000217, A000567, A046184.

Programs

Magma

I:=[0,14]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+12: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016

Mathematica

RecurrenceTable[{a[1] == 0, a[2] == 14, a[n] == 14 a[n-1]- a[n-2] + 12}, a, {n, 20}] (* Vincenzo Librandi, Mar 05 2016 *)

PARI

concat(0, Vec(2*x^2*(x-7) / ((x-1)*(x^2-14*x+1)) + O(x^100)))

Formula

a(n) = 15*a(n-1)-15*a(n-2)+a(n-3).

G.f.: 2*x^2*(x-7) / ((x-1)*(x^2-14*x+1)).

a(n) = (-6-(7-4*sqrt(3))^n*(3+2*sqrt(3))+(-3+2*sqrt(3))*(7+4*sqrt(3))^n)/6. - Colin Barker, Mar 05 2016

a(n) = 14*a(n-1) - a(n-2) + 12. - Vincenzo Librandi, Mar 05 2016