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%I #12 Sep 11 2013 03:42:04
%S 1,0,1,1,3,7,16,45,110,308,819,2275,6328,17748,50388,143412,411939,
%T 1187329,3441559,10015005,29255655,85766655,252201690,743819115,
%U 2199446652,6519727800,19369551936,57665571072,172011364452,514021640564,1538650042952
%N Series reversion of x-x^3-x^4.
%H Vincenzo Librandi, <a href="/A217358/b217358.txt">Table of n, a(n) for n = 1..1000</a>
%F Conjecture: 46*n*(n-1)*(n-2)*a(n) -(n-1)*(n-2)*(11*n-74)*a(n-1) -(n-2)*(336*n^2-1359*n+1351)*a(n-2) +(-347*n^3+2190*n^2-3861*n+1330)*a(n-3) + 8*(2*n-7)*(4*n-15)*(4*n-17)*a(n-4) = 0.
%F Recurrence (order 3): 23*(n-2)*(n-1)*n*(9*n-25)*a(n) = -(n-2)*(n-1)*(54*n^2 - 231*n + 248)*a(n-1) + (n-2)*(1485*n^3 - 10065*n^2 + 22292*n - 16088)*a(n-2) + 8*(2*n-5)*(4*n-13)*(4*n-11)*(9*n-16)*a(n-3). - _Vaclav Kotesovec_, Sep 10 2013
%F a(n) ~ c*d^n/n^(3/2), where d = 3/23*(2367+966*sqrt(3))^(1/3)+423/(23*(2367+966*sqrt(3))^(1/3))-2/23 = 3.145200906807902443... is the root of the equation -256 - 165*d + 6*d^2 + 23*d^3 = 0 and c = 1/48*sqrt(2)*sqrt((80793 + 65184*sqrt(3))^(1/3)*((80793 + 65184 * sqrt(3))^(2/3)-1839+9*(80793 + 65184 * sqrt(3))^(1/3)))/((80793 + 65184 * sqrt(3))^(1/3)*sqrt(Pi)) = 0.098446219937815765... - _Vaclav Kotesovec_, Sep 10 2013
%e If y= x-x^3-x^4, then x= y + y^3 + y^4 +3*y^5 +7*y^6 +16*y^7 + ...
%t Rest[CoefficientList[InverseSeries[Series[x - x^3 - x^4, {x, 0, 20}], x], x]] (* _Vaclav Kotesovec_, Sep 10 2013 *)
%Y Cf. A049140 (reversion of x-x^2-x^4).
%K nonn
%O 1,5
%A _R. J. Mathar_, Oct 01 2012