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A103737
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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).
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2
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0, 0, 3, 7, 76, 164, 1679, 3611, 36872, 79288, 809515, 1740735, 17772468, 38216892, 390184791, 839030899, 8566292944, 18420462896, 188068259987, 404411152823, 4128935426780, 8878624899220, 90648511129183, 194925336630027, 1990138309415256, 4279478780961384, 43692394296006459
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OFFSET
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1,3
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COMMENTS
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By construction and recurrence, 30*a(n)^2 + 30*a(n) + 1 = j(n)^2.
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LINKS
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FORMULA
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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
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MATHEMATICA
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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 *)
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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