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A251867 Numbers n such that n^2 + (n+1)^2 is equal to the sum of the hexagonal numbers H(m) and H(m+1) for some m.

%I A251867 #23 Sep 08 2022 08:46:10
%S A251867 0,14,492,16730,568344,19306982,655869060,22280241074,756872327472,
%T A251867 25711378892990,873430010034204,29670908962269962,1007937474707144520,
%U A251867 34240203231080643734,1163158972382034742452,39513164857758100599650,1342284446191393385645664
%N A251867 Numbers n such that n^2 + (n+1)^2 is equal to the sum of the hexagonal numbers H(m) and H(m+1) for some m.
%C A251867 Also nonnegative integers y in the solutions to 4*x^2-2*y^2+2*x-2*y = 0, the corresponding values of x being A220185.
%H A251867 Colin Barker, <a href="/A251867/b251867.txt">Table of n, a(n) for n = 1..654</a>
%H A251867 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).
%F A251867 a(n) = 35*a(n-1)-35*a(n-2)+a(n-3).
%F A251867 G.f.: -2*x^2*(x+7) / ((x-1)*(x^2-34*x+1)).
%F A251867 a(n) = A220185(n) + A001542(n-1)^2. - _Alexander Samokrutov_, Sep 05 2015
%F A251867 a(n) = (-4+(10+7*sqrt(2))*(17+12*sqrt(2))^(-n)+(10-7*sqrt(2))*(17+12*sqrt(2))^n)/8. - _Colin Barker_, Mar 02 2016
%F A251867 a(n) = A105635(4*n-4). - _Greg Dresden_, Aug 30 2021
%e A251867 14 is in the sequence because 14^2+15^2 = 196+225 = 421 = 190+231 = H(10)+H(11).
%t A251867 LinearRecurrence[{35, -35, 1}, {0, 14, 492}, 20] (* _Vincenzo Librandi_, Sep 06 2015 *)
%o A251867 (PARI) concat(0, Vec(-2*x^2*(x+7)/((x-1)*(x^2-34*x+1)) + O(x^100)))
%o A251867 (Magma) I:=[0,14,492]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // _Vincenzo Librandi_, Sep 06 2015
%Y A251867 Cf. A000290, A000384, A220185, A105635.
%K A251867 nonn,easy
%O A251867 1,2
%A A251867 _Colin Barker_, Dec 10 2014