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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.

%I A251963 #15 Sep 08 2022 08:46:10
%S A251963 0,14,208,2910,40544,564718,7865520,109552574,1525870528,21252634830,
%T A251963 296011017104,4122901604638,57424611447840,799821658665134,
%U A251963 11140078609864048,155161278879431550,2161117825702177664,30100488280951055758,419245718107612602960
%N 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.
%C A251963 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 A251963 Colin Barker, <a href="/A251963/b251963.txt">Table of n, a(n) for n = 1..874</a>
%H A251963 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).
%F A251963 a(n) = 15*a(n-1)-15*a(n-2)+a(n-3).
%F A251963 G.f.: 2*x^2*(x-7) / ((x-1)*(x^2-14*x+1)).
%F A251963 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 A251963 a(n) = 14*a(n-1) - a(n-2) + 12. - _Vincenzo Librandi_, Mar 05 2016
%e A251963 14 is in the sequence because T(14)+T(15) = 105+120 = 225 = N(9).
%t A251963 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 A251963 (PARI) concat(0, Vec(2*x^2*(x-7) / ((x-1)*(x^2-14*x+1)) + O(x^100)))
%o A251963 (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 A251963 Cf. A000217, A000567, A046184.
%K A251963 nonn,easy
%O A251963 1,2
%A A251963 _Colin Barker_, Dec 11 2014