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%I #10 Feb 13 2023 19:24:50
%S 0,0,0,0,1,2,-3,5,7,-9,12,15,-18,22,26,-30,35,40,-45,51,57,-63,70,77,
%T -84,92,100,-108,117,126,-135,145,155,-165,176,187,-198,210,222,-234,
%U 247,260,-273,287,301,-315,330,345,-360,376,392,-408,425,442,-459,477
%N A000332(n) = a(n)*(3*a(n) - 1)/2.
%C As the formula in the description shows, all members of A000332 belong to the generalized pentagonal sequence (A001318). A001318 also lists all nonnegative numbers that belong to A145919.
%H G. C. Greubel, <a href="/A145919/b145919.txt">Table of n, a(n) for n = 0..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PentagonalNumber.html">Pentagonal Number</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PentatopeNumber.html">Pentatope Number</a>.
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,3,0,0,-3,0,0,1).
%F a(n+3) = A001840(n) when 3 does not divide n, A001840(n)*-1 otherwise.
%F After first two zeros, this sequence consists of all values of A001318(n) and A045943(n)*(-1), n>=0, sorted in order of increasing absolute value.
%F G.f.: (-x^4*(x^4+2*x^3-3*x^2+2*x+1))/((x-1)^3*(1+x^2+x)^3). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
%e a(6) = -3 and A000332(6) = (-3)(-10)/2 = 15.
%t CoefficientList[Series[(-x^4*(x^4 + 2*x^3 - 3*x^2 + 2*x + 1))/((x - 1)^3*(1 + x^2 + x)^3), {x,0,50}], x] (* _G. C. Greubel_, Jun 13 2017 *)
%t LinearRecurrence[{0,0,3,0,0,-3,0,0,1},{0,0,0,0,1,2,-3,5,7},60] (* _Harvey P. Dale_, Feb 13 2023 *)
%o (PARI) x='x+O('x^50); concat([0, 0, 0, 0], Vec((-x^4*(x^4 +2*x^3 -3*x^2 +2*x +1))/((x-1)^3*(1+x^2+x)^3))) \\ _G. C. Greubel_, Jun 13 2017
%Y Cf. A000326, A145920.
%K easy,sign
%O 0,6
%A _Matthew Vandermast_, Oct 28 2008
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