%I #15 Sep 08 2022 08:46:10

%S 0,14,208,2910,40544,564718,7865520,109552574,1525870528,21252634830,

%T 296011017104,4122901604638,57424611447840,799821658665134,

%U 11140078609864048,155161278879431550,2161117825702177664,30100488280951055758,419245718107612602960

%N 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.

%C 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.

%H Colin Barker, <a href="/A251963/b251963.txt">Table of n, a(n) for n = 1..874</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).

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

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

%F 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

%F a(n) = 14*a(n-1) - a(n-2) + 12. - _Vincenzo Librandi_, Mar 05 2016

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

%t 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 *)

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

%o (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

%Y Cf. A000217, A000567, A046184.

%K nonn,easy

%O 1,2

%A _Colin Barker_, Dec 11 2014