A235347 Series reversion of x*(1-3*x^2)/(1-x^2) in odd-order powers.

1, 2, 14, 130, 1382, 15906, 192894, 2427522, 31405430, 415086658, 5580629870, 76080887042, 1049295082630, 14613980359010, 205246677882078, 2903566870820610, 41337029956899222, 591796707042765954, 8514525059135909070, 123048063153362454402

Offset: 0

Author

Fung Lam, Jan 10 2014

Keywords

• nonn
• easy

Comments

This sequence is implied in the solutions of magnetohydrodynamics equations in R^3 for incompressible, electrically-conducting fluids in the presence of a strong Lorentz force. a(n) = numbers of allowable magneto-vortical eddies in terms of initial conditions.

Links

• Georg Fischer, Table of n, a(n) for n = 0..1000 (first 940 terms from _Fung Lam_)
• Fung Lam, On the Well-posedness of Magnetohydrodynamics Equations for Incompressible Electrically-Conducting Fluids, arXiv:1401.2029 [physics.flu-dyn], 2014-2023.

Crossrefs

Cf. A027307, A107841, A235352 (same except for signs).

Programs

Maple

Order := 60 ;
solve(series(x*(1-3*x^2)/(1-x^2),x)=y,x) ;
convert(%,polynom) ;
seq(coeff(%,y,2*i+1),i=0..Order/2) ; # R. J. Mathar, Jul 20 2023

Mathematica

Table[(CoefficientList[InverseSeries[Series[x*(1-3*x^2)/(1-x^2),{x,0,40}],x],x])[[n]],{n,2,40,2}] (* Vaclav Kotesovec, Jan 29 2014 *)

PARI

v=Vec( serreverse(x*(1-3*x^2)/(1-x^2) +O(x^66) ) ); vector(#v\2,j,v[2*j-1]) \\ Joerg Arndt, Jan 14 2014

Formula

G.f.: (exp(4*Pi*i/3)*u + exp(2*Pi*i/3)*v + x/9)/x, where i=sqrt(-1),

u = (1/9)*(x^3 - 108 *x + 9*sqrt(-9 + 141*x^2 - 3*x^4))^(1/3), and

v = (1/9)*(x^3 - 108 *x - 9*sqrt(-9 + 141*x^2 - 3*x^4))^(1/3).

a(n) = [x^n] 2*Sum_{j = 1..n} ((Sum_{k = 1..n} a(k)*x^(2*k-1))^(2*j+1)), a(1) = 1, with offset by 1.

D-finite with recurrence 12*n*(2*n+1)*a(n) +(-382*n^2+391*n-90)*a(n-1) +3*(34*n^2-132*n+125)*a(n-2) -(2*n-5)*(n-3)*a(n-3)=0. - R. J. Mathar, Mar 24 2023

From Seiichi Manyama, Aug 09 2023: (Start)

a(n) = (-1)^n * Sum_{k=0..n} (-3)^k * binomial(n,k) * binomial(2*n+k+1,n) / (2*n+k+1).

a(n) = (1/n) * Sum_{k=0..n-1} 2^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-k) for n > 0.

a(n) = (1/n) * Sum_{k=1..n} 2^k * 3^(n-k) * binomial(n,k) * binomial(2*n,k-1) for n > 0. (End)

From Peter Bala, Sep 08 2024: (Start)

a(n) = 2*Jacobi_P(n-1, 1, n+1, 5)/n for n >= 1.

Second-order recurrence: 3*n*(2*n + 1)*(13*n - 17)*a(n) = (1222*n^3 - 2820*n^2 + 1877*n - 360)*a(n-1) - (n - 2)*(13*n - 4)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 2. (End)