A317135 - OEIS

%I #8 Jul 22 2018 03:41:54

%S 1,1,4,18,92,504,2897,17235,105233,655687,4152461,26650675,172961053,

%T 1133152365,7484233381,49780558057,333155274455,2241793462963,

%U 15158132783457,102938343190333,701783155862751,4801340686195787,32954688422181493,226853921031126233,1565828056187052419,10834714749540715871,75142241780769151970,522240807596491906516

%N G.f.: Sum_{n>=0} binomial(4*n+2, n)/(2*n+1) * x^(n+1)/(1+x)^(2*n+1).

%C Equals the self-convolution square-root of A317134.

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

%F (1) A(x) = ( (1 + x*A(x)^2)^2 + sqrt( (1 + x*A(x)^2)^4 - 4*x*A(x)^2 ) ) / 2.

%F (2) A(x) = sqrt( (1/x) * Series_Reversion( 4*x/((1+x)^2 + sqrt( (1+x)^4 - 4*x ))^2 ) ).

%F (3) A(x) = Sum_{n>=0} binomial(4*n+2, n)/(2*n+1) * x^(n+1)/(1+x)^(2*n+1).

%F a(n) ~ 37^(1/4) * (101 + 16*sqrt(37))^(n + 1/2) / (4*sqrt(Pi) * n^(3/2) * 3^(3*n + 5/2)). - _Vaclav Kotesovec_, Jul 22 2018

%e G.f.: A(x) = 1 + x + 4*x^2 + 18*x^3 + 92*x^4 + 504*x^5 + 2897*x^6 + 17235*x^7 + 105233*x^8 + 655687*x^9 + 4152461*x^10 + ...

%e such that

%e A(x) = 1/(1+x) + 2*x/(1+x)^3 + 9*x^2/(1+x)^5 + 52*x^3/(1+x)^7 + 340*x^4/(1+x)^9 + 2394*x^5/(1+x)^11 + ... + A069271(n)*x^n/(1+x)^(2*n+1) + ...

%t CoefficientList[Sqrt[1/x * InverseSeries[Series[4*x/((1 + x)^2 + Sqrt[(1 + x)^4 - 4*x])^2, {x, 0, 30}], x]], x] (* _Vaclav Kotesovec_, Jul 22 2018 *)

%o (PARI) {a(n) = my(A = sum(m=0, n, binomial(4*m+2, m)/(2*m+1) * x^m / (1+x +x*O(x^n))^(2*m+1))); polcoeff(A, n)}

%o for(n=0, 30, print1(a(n), ", "))

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

%o for(n=0, 30, print1(a(n), ", "))

%Y Cf. A317134.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jul 22 2018