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%I #15 Jan 17 2024 09:06:51
%S 5,8002,11635629,16918197290,24599047224757,35766997746600114,
%T 52005190124509341725,75615510674038836268762,
%U 109944900514862343425438949,159859809733099173301751963810,232436053407025683118403929941517,337961861794005610154986012383002634
%N Numbers x such that there exists n in N with (x+1)^3-x^3=91*n^2.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1455,-1455,1).
%F a(n+2) = 1454*a(n+1)-a(n)+726.
%F G.f.: x*(6*x^2-727*x-5) / ((x-1)*(x^2-1454*x+1)). - _Colin Barker_, Oct 18 2014
%e a(1)=5 because the first relation is (5+1)^3-5^3=91*1^2.
%t LinearRecurrence[{1455, -1455, 1}, {5, 8002, 11635629}, 15] (* _Paolo Xausa_, Jan 17 2024 *)
%o (PARI) Vec(x*(6*x^2-727*x-5)/((x-1)*(x^2-1454*x+1)) + O(x^20)) \\ _Colin Barker_, Oct 18 2014
%K easy,nonn
%O 1,1
%A _Richard Choulet_, Oct 12 2008, Oct 13 2008
%E Editing and more terms from _Colin Barker_, Oct 18 2014
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