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%I #7 Jun 13 2015 00:55:19
%S 0,2,68,286,6760,28122,662508,2755766,64919120,270037042,6361411348,
%T 26460874446,623353393080,2592895658762,61082271110588,
%U 254077313684326,5985439215444640,24896983845405282,586511960842464228,2439650339536033406
%N Numbers n such that the sum of the triangular numbers T(n) and T(n+1) is equal to the sum of the octagonal numbers N(m) and N(m+1) for some m.
%C Also nonnegative integers y in the solutions to 6*x^2-y^2+2*x-2*y = 0, the corresponding values of x being A220755.
%H Colin Barker, <a href="/A251793/b251793.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,98,-98,-1,1).
%F a(n) = a(n-1)+98*a(n-2)-98*a(n-3)-a(n-4)+a(n-5).
%F G.f.: -2*x^2*(3*x^3+11*x^2+33*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)).
%e 68 is in the sequence because T(68)+T(69) = 2346+2415 = 4761 = 2296+2465 = N(28)+N(29).
%o (PARI) concat(0, Vec(-2*x^2*(3*x^3+11*x^2+33*x+1)/((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100)))
%Y Cf. A000217, A000567, A220755.
%K nonn,easy
%O 1,2
%A _Colin Barker_, Dec 09 2014
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