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A103737 Define a(1)=0, a(2)=0, a(3)=3, a(4)=7 such that from i=1 to 4: 30*a(i)^2 + 30*a(i) + 1 = j(i)^2, j(1)=1, j(2)=1, j(3)=19, j(4)=41 Then a(n) = a(n-4) + 4*sqrt(30*(a(n-2)^2) + 30*a(n-2) + 1).

%I A103737 #14 Mar 05 2025 18:02:47
%S A103737 0,0,3,7,76,164,1679,3611,36872,79288,809515,1740735,17772468,
%T A103737 38216892,390184791,839030899,8566292944,18420462896,188068259987,
%U A103737 404411152823,4128935426780,8878624899220,90648511129183,194925336630027,1990138309415256,4279478780961384,43692394296006459
%N A103737 Define a(1)=0, a(2)=0, a(3)=3, a(4)=7 such that from i=1 to 4: 30*a(i)^2 + 30*a(i) + 1 = j(i)^2, j(1)=1, j(2)=1, j(3)=19, j(4)=41 Then a(n) = a(n-4) + 4*sqrt(30*(a(n-2)^2) + 30*a(n-2) + 1).
%C A103737 By construction and recurrence, 30*a(n)^2 + 30*a(n) + 1 = j(n)^2.
%H A103737 G. C. Greubel, <a href="/A103737/b103737.txt">Table of n, a(n) for n = 1..1000</a>
%H A103737 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,22,-22,-1,1).
%F A103737 G.f.: x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
%t A103737 Rest[CoefficientList[Series[x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)), {x, 0, 50}], x]] (* _G. C. Greubel_, Jul 15 2018 *)
%t A103737 LinearRecurrence[{1,22,-22,-1,1},{0,0,3,7,76},40] (* _Harvey P. Dale_, Mar 05 2025 *)
%o A103737 (PARI) x='x+O('x^30); concat([0,0], Vec(x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)))) \\ _G. C. Greubel_, Jul 15 2018
%o A103737 (Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!(x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)))); // _G. C. Greubel_, Jul 15 2018
%Y A103737 Cf. A053141, A103200.
%K A103737 nonn
%O A103737 1,3
%A A103737 _Pierre CAMI_, Mar 27 2005
%E A103737 Terms a(19) onward added by _G. C. Greubel_, Jul 15 2018