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Series reversion of (sqrt(1+4*x) - 1)/2 - x^3.
2

%I #23 Jan 30 2020 21:29:17

%S 1,1,1,5,12,35,122,390,1320,4631,16185,57707,208348,756840,2775012,

%T 10246206,38043339,142045387,532888840,2007554241,7592537590,

%U 28814794105,109702322730,418868083725,1603584623355,6154156653687,23671591739306,91242219125712,352378515196920,1363360128627380

%N Series reversion of (sqrt(1+4*x) - 1)/2 - x^3.

%H Robert Israel, <a href="/A229043/b229043.txt">Table of n, a(n) for n = 1..1647</a>

%H Robert Israel, <a href="/A229043/a229043.pdf">Recurrence</a>

%F G.f. A(x) satisfies:

%F (1) A(x)^3 = Sum_{n>=1} d^(n-1)/dx^(n-1) x^(3*n)*(1+x)^(3*n)/n! ).

%F (2) A(x)^3 = A(x)*C(-A(x)) - x^3, where C(x) = 1 + x*C(x)^2 is the g.f. of A000108.

%F (3) A(x) = G(x)*(1 + G(x)) where G(x) = x + A(x)^3.

%F a(n) ~ r^(1/2-n) / (2 * sqrt(3*Pi*s*(1+9*s^5)) * n^(3/2)), where s = 0.44683030245197... is the root of the equation 9*s^4*(1+4*s) = 1 and r = -1/2 + 1/(6*s^2) - s^3 = 0.24555068038... - _Vaclav Kotesovec_, Jan 22 2014

%F D-finite with recurrence: Sequence satisfies a 9th-order linear recurrence with coefficients that are polynomials in n of degree 5: see link. - _Robert Israel_, May 14 2018

%e G.f.: A(x) = x + x^2 + x^3 + 5*x^4 + 12*x^5 + 35*x^6 + 122*x^7 + 390*x^8 +...

%e where the series reversion of g.f. A(x) begins:

%e (sqrt(1+4*x) - 1)/2 - x^3 = x - x^2 + x^3 - 5*x^4 + 14*x^5 - 42*x^6 + 132*x^7 - 429*x^8 +...+ (-1)^(n-1)*A000108(n-1)*x^n +...

%e The cube of the g.f. equals the series:

%e A(x)^3 = x^3*(1+x)^3 + d/dx x^6*(1+x)^6/2! + d^2/dx^2 x^9*(1+x)^9/3! + d^3/dx^3 x^12*(1+x)^12/4! + d^4/dx^4 x^15*(1+x)^15/5! +...

%e A(x)^3 = x^3 + 3*x^4 + 6*x^5 + 22*x^6 + 72*x^7 + 225*x^8 + 790*x^9 + 2739*x^10 +...

%e RELATED EXPANSIONS.

%e G.f. A(x) = G(x)*(1 + G(x)) = (G(x) - x)^(1/3) where G(x) begins:

%e G(x) = x + x^3 + 3*x^4 + 6*x^5 + 22*x^6 + 72*x^7 + 225*x^8 + 790*x^9 +...

%p with(gfun):

%p S:= solve((sqrt(1+4*x)-1)/2-x^3=y, x):

%p DE:= holexprtodiffeq(S,g(y)):

%p Rec:= diffeqtorec(DE,g(y),a(n)):

%p f:= rectoproc(Rec,a(n),remember):

%p map(f, [$1..40]); # _Robert Israel_, May 14 2018

%t Rest[CoefficientList[InverseSeries[Series[(Sqrt[1+4*x]-1)/2-x^3, {x, 0, 20}], x],x]] (* _Vaclav Kotesovec_, Jan 22 2014 *)

%o (PARI) {a(n)=polcoeff( serreverse( (sqrt(1+4*x +x*O(x^n)) - 1)/2 - x^3 ), n)}

%o for(n=1, 30, print1(a(n), ", "))

%o (PARI) /* G.f. A(x) = sqrt(G(x) - x) where G(x) = x + G(x)^3*(1 + G(x))^3 */

%o {a(n)=local(G=serreverse(x-x^3*(1+x)^3+x^3*O(x^n)));polcoeff((G-x)^(1/3),n)}

%o for(n=1, 30, print1(a(n), ", "))

%o (PARI) {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D} \\ n-th derivative

%o {a(n)=local(A3=x); A3=sum(m=1, n, Dx(m-1, x^(3*m)*(1+x+x*O(x^n))^(3*m)/m!)); polcoeff(A3^(1/3), n)}

%o for(n=1, 30, print1(a(n), ", "))

%o (PARI) x='x+O('x^99); Vec(serreverse(((1+4*x)^(1/2)-1)/2-x^3)) \\ _Altug Alkan_, May 14 2018

%Y Cf. A229042.

%K nonn

%O 1,4

%A _Paul D. Hanna_, Nov 02 2013

%E Offset corrected by _Vaclav Kotesovec_, Jan 22 2014