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A115344
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Numerators of asymptotic expansion of first root of Ziegler's cubic in an imaginary quadratic field.
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2
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-1, 1, -1, 3, -6, 18, -45, 136, -378, 1156, -3405, 10549, -32175, 100915, -314834, 998323, -3163683, 10127020, -32462265, 104751043, -338742887, 1100559573, -3583933846, 11711868458, -38358103030, 125974533997, -414566089320, 1367353737806, -4518185596293
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OFFSET
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0,4
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LINKS
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FORMULA
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G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) x = -(1+x)*A(x) - A(x)^2 + x*A(x)^3. - Paul D. Hanna, May 30 2014
(2) x = -A(x)*(1 + A(x)) / (1 + A(x) - A(x)^3). - Paul D. Hanna, May 31 2014
(3) A(x) = -x/Series_Reversion(x*(1 - Series_Reversion(x/(1 - 2*x + 3*x^2 - x^3)))). - Paul D. Hanna, May 31 2014
Recurrence: n*(n+1)*(28*n^2 - 94*n + 51)*a(n) = -4*n*(14*n^3 - 54*n^2 + 73*n - 48)*a(n-1) + (n-3)*(140*n^3 - 330*n^2 + 19*n + 216)*a(n-2) + 6*(n-3)*(28*n^3 - 108*n^2 + 57*n + 118)*a(n-3) + 23*(n-4)*(n-3)*(28*n^2 - 38*n - 15)*a(n-4). - Vaclav Kotesovec, May 30 2014
a(n) ~ (-1)^(n+1) * sqrt(s*(s-1)/(3*r*s-1)) / (2*sqrt(Pi) * n^(3/2)* r^n), where r = 2/(1+sqrt(13+16*sqrt(2))) = 0.2869905464691794898..., s = 1/2 + 1/sqrt(2) + 1/2*sqrt(2*sqrt(2)-1) = 1.88320350591352586... . - Vaclav Kotesovec, May 30 2014
a(n) = Sum_{k=0..n}(binomial(n,k)*Sum_{i=0..n-k-1}(2^(k-i)*binomial(k,i)*(-1)^(i+k)*binomial(2*n-i-2*k-2,n-k-1)))/n, n>0, a(0)=-1. - Vladimir Kruchinin, Mar 15 2016
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EXAMPLE
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-1 + 1/t - 1/t^2 + 3/t^3 - 6/t^4 + 18/t^5 - 45/t^6 + 136/t^7 - 378/t^8...
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MATHEMATICA
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nmax=30; aa=ConstantArray[0, nmax]; aa[[1]]=1; Do[AGF=-1+Sum[aa[[n]]*x^n, {n, 1, j-1}]+koef*x^j; sol=Solve[Coefficient[-(1+x)*AGF-AGF^2+x*AGF^3-x, x, j]==0, koef][[1]]; aa[[j]]=koef/.sol[[1]], {j, 2, nmax}]; Flatten[{-1, aa}] (* Vaclav Kotesovec, May 30 2014 *)
CoefficientList[-x/InverseSeries[x*(1-InverseSeries[Series[x/(1-2*x+3*x^2-x^3), {x, 0, 20}], x]), x], x] (* Vaclav Kotesovec, May 31 2014 after Paul D. Hanna *)
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PROG
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(PARI) {a(n)=polcoeff(-x/serreverse(x*(1-serreverse(x/(1 - 2*x + 3*x^2 - x^3 +x*O(x^n))))), n)}
(Maxima)
a(n):=if n=0 then -1 else sum(binomial(n, k)*sum(2^(k-i)*binomial(k, i)*(-1)^(i+k)*binomial(2*n-i-2*k-2, n-k-1), i, 0, n-k-1), k, 0, n)/n; /* Vladimir Kruchinin, Mar 15 2016 */
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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