%I #16 Jan 30 2020 21:29:15
%S 1,5,35,250,1795,12910,92910,668820,4815075,34667110,249598330,
%T 1797091180,12938997710,93160575500,670755400700,4829436210600,
%U 34771931021475,250357867996950,1802576519933250,12978550465880700,93445561587077850,672808036862840100,4844217840946181700
%N Expansion of (1+sqrt(1-4*x))/(6*sqrt(1-4*x)-4).
%C Apply the Riordan matrix ((1+sqrt(1-4x))/2,(1-sqrt(1-4x))/2) (inverse (1/(1-x),x(1-x))) to 6^n. In general, (1+sqrt(1-4x))(k*sqrt(1-4x)-(k-2)) results from applying the Riordan matrix ((1+sqrt(1-4x))/2,(1-sqrt(1-4x))/2) (inverse of (1/(1-x),x(1-x))) to k^n. We then have a(n)=0^n+sum{i=0..n, (k-1)^(i+1)*C(2n-1,n-i-1)2(i+1)/(n+i+1)}.
%H Vincenzo Librandi, <a href="/A104532/b104532.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) = 0^n + sum{k=0..n, 5^(k+1)*C(2n-1, n-k-1)*2*(k+1)/(n+k+1)}.
%F D-finite with recurrence: 5*n*a(n) = 2*(28*n-15)*a(n-1) - 72*(2*n-3)*a(n-2). - _Vaclav Kotesovec_, Oct 17 2012
%F a(n) ~ 2^(2*n+1)*3^(2*n-1)/5^n. - _Vaclav Kotesovec_, Oct 17 2012
%t CoefficientList[Series[(1+Sqrt[1-4x])/(6Sqrt[1-4x]-4), {x,0,20}], x] (* _Harvey P. Dale_, Apr 11 2011 *)
%o (PARI) x='x+O('x^66); Vec((1+sqrt(1-4*x))/(6*sqrt(1-4*x)-4)) \\ _Joerg Arndt_, May 13 2013
%Y Cf. A088218, A067336, A076025, 104530, 104531.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Mar 12 2005