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%I #23 Jan 30 2020 21:29:17
%S 1,2,26,476,10150,236060,5807076,148581048,3913759878,105424703020,
%T 2890693930124,80413849328904,2263896023453532,64381391412987672,
%U 1846729385267277960,53367451809002583408,1552274439636853988550,45408989873571191613900,1335107241077282661195900
%N G.f.: ( (1 - sqrt(1-32*x)) / (16*x) )^(1/4).
%H G. C. Greubel, <a href="/A228411/b228411.txt">Table of n, a(n) for n = 0..665</a>
%F G.f. A(x) satisfies:
%F (1) A(x) = exp( x*A(x)^8 + Integral(A(x)^8 dx) ).
%F (2) A(x)^4 = 1 + 8*x*A(x)^8, thus A(x) = C(8*x)^(1/4) where C(x) is the Catalan function (A000108).
%F a(n) ~ 2^(5*n-3+1/4)/(sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Nov 10 2013
%F D-finite with recurrence: n*(4*n+1)*a(n) -2*(8*n-3)*(8*n-7)*a(n-1)=0. - _R. J. Mathar_, Oct 08 2016
%F a(n) = 8^n*binomial(2*n + 1/4, n)/(8*n + 1). - _Vladimir Reshetnikov_, Oct 12 2016
%e G.f.: A(x) = 1 + 2*x + 26*x^2 + 476*x^3 + 10150*x^4 + 236060*x^5 +...
%e where
%e A(x)^4 = 1 + 8*x + 128*x^2 + 2560*x^3 + 57344*x^4 + 1376256*x^5 +...+ A000108(n)*8^n*x^n +...
%t CoefficientList[Series[((1-Sqrt[1-32*x])/(16*x))^(1/4), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Nov 10 2013 *)
%t Table[8^n Binomial[2 n + 1/4, n]/(8 n + 1), {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 12 2016 *)
%o (PARI) /* G.f.: ( (1 - sqrt(1-32*x)) / (16*x) )^(1/4): */
%o {a(n)=polcoeff(( (1 - sqrt(1-32*x +x^2*O(x^n))) / (16*x) )^(1/4),n)}
%o for(n=0,30,print1(a(n),", "))
%o (PARI) /* G.f.: A(x) = C(8*x)^(1/4), C(x) is Catalan function: */
%o {a(n)=polcoeff((serreverse(x-8*x^2 +x^2*O(x^n))/x)^(1/4),n)}
%o for(n=0,30,print1(a(n),", "))
%o (PARI) /* G.f.: A(x) = exp( x*A(x)^8 + Integral(A(x)^8 dx) ): */
%o {a(n)=local(A=1+x);for(i=1,n,A=exp(x*A^8+intformal(A^8+x*O(x^n))));polcoeff(A,n)}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A000108, A159318.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Nov 09 2013
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