%I #12 Sep 08 2022 08:45:17
%S 0,0,3,7,76,164,1679,3611,36872,79288,809515,1740735,17772468,
%T 38216892,390184791,839030899,8566292944,18420462896,188068259987,
%U 404411152823,4128935426780,8878624899220,90648511129183,194925336630027,1990138309415256,4279478780961384,43692394296006459
%N 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 By construction and recurrence, 30*a(n)^2 + 30*a(n) + 1 = j(n)^2.
%H G. C. Greubel, <a href="/A103737/b103737.txt">Table of n, a(n) for n = 1..1000</a>
%F 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 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 *)
%o (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 (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 Cf. A053141, A103200.
%K nonn
%O 1,3
%A _Pierre CAMI_, Mar 27 2005
%E Terms a(19) onward added by _G. C. Greubel_, Jul 15 2018