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A235350
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Series reversion of x*(1-2*x-x^2)/(1-x^2).
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1
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1, 2, 8, 42, 248, 1570, 10416, 71474, 503088, 3612226, 26353720, 194806458, 1455874792, 10982013250, 83504148192, 639360351074, 4925190101600, 38144591091970, 296837838901992, 2319880586624714, 18200693844341720, 143294043656426082, 1131747417739664528
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OFFSET
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1,2
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COMMENTS
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Derived series from A107841. The reversion has a quadratic power in x in the denominator. The general form reads x*(1-p*x-q*x^2)/(1-q*x^2).
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LINKS
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FORMULA
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G.f.: (exp(4*Pi*i/3)*u + exp(2*Pi*i/3)*v - 2/3)/x, where i=sqrt(-1),
u = 1/3*(-17+3*x-6*x^2+x^3+3*sqrt(-6+54*x-30*x^2+18*x^3-3*x^4))^(1/3), and
v = 1/3*(-17+3*x-6*x^2+x^3-3*sqrt(-6+54*x-30*x^2+18*x^3-3*x^4))^(1/3).
First few terms can be obtained by Maclaurin's expansion of G.f.
D-finite with recurrence 6*n*(n-1)*a(n) -(n-1)*(52*n-75)*a(n-1) +(2*n+3)*(5*n-11)*a(n-2) +2*(5*n^2-62*n+150)*a(n-3) +(-13*n^2+130*n-321)*a(n-4) +(7*n-37)*(n-6)*a(n-5) -(n-6)*(n-7)*a(n-6)=0. - R. J. Mathar, Mar 24 2023
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MATHEMATICA
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Rest[CoefficientList[InverseSeries[Series[x*(1-2*x-x^2)/(1-x^2), {x, 0, 20}], x], x]] (* Vaclav Kotesovec, Jan 29 2014 *)
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PROG
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(Python)
# a235350. The list a has been calculated (len(a)>=3).
m = len(a)
d = 0
for i in range (1, m+3):
....for j in range (1, m+3):
........if (i+j)%m ==0 and (i+j) <= m:
............d = d + a[i-1]*a[j-1]
f = 0
for i in range (1, m+1):
....for j in range (1, m+1):
........if (i+j)%(m+1) ==0 and (i+j) <= (m+1):
............f = f + a[i-1]*a[j-1]
g = 0
for i in range (1, m+1):
....for j in range (1, m+1):
........for k in range (1, ip):
............if (i+j+k)%(m+1) ==0 and (i+j+k) <= (m+1):
................g = g + a[i-1]*a[j-1]*a[k-1]
y = g + 2*f - d
# a235350.
(PARI) Vec(serreverse(x*(1-2*x-x^2)/(1-x^2)+O(x^66))) \\ Joerg Arndt, Jan 17 2014
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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