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%I #12 Feb 23 2014 21:05:49
%S 1,1,5,45,605,11045,257005,7288245,243870205,9401560645,410141056205,
%T 19966451812245,1072718714991005,63033317759267045,
%U 4020725747388170605,276661592017425909045,20424931173615717011005
%N INVERT transform (with offset) of quintuple factorials (A008546), where g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^5]/A(x)^5.
%H Vincenzo Librandi, <a href="/A112940/b112940.txt">Table of n, a(n) for n = 0..200</a>
%F G.f. satisfies: A(x) = 1+x + 5*x^2*[d/dx A(x)]/A(x) (log derivative). G.f.: A(x) = 1+x +5*x^2/(1-9*x -5*2*4*x^2/(1-19*x -5*3*9*x^2/(1-29*x -5*4*13*x^2/(1-39*x -... -5*n*(5*n-6)*x^2/(1-(10*n-1)*x -...)))) (continued fraction). G.f.: A(x) = 1/(1-1*x/(1 -4*x/(1-5*x/(1 -9*x/(1-10*x/(1 -14*x/(1-15*x/(1 -...)))))))) (continued fraction).
%F G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 - x*(5*k+4)/(1 - x*(5*k+5)/Q(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Mar 20 2013
%F a(n) ~ (n-1)! * 5^(n-1) / (GAMMA(4/5) * n^(1/5)). - _Vaclav Kotesovec_, Feb 22 2014
%e A(x) = 1 + x + 5*x^2 + 45*x^3 + 605*x^4 + 11045*x^5 +...
%e 1/A(x) = 1 - x - 4*x^2 - 36*x^3 - 504*x^4 -... -A008546(n)*x^(n+1) -...
%t CoefficientList[Series[1/(1 + 1/5*ExpIntegralE[4/5, -1/(5*x)]/E^(1/(5*x))), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 22 2014 *)
%o (PARI) {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+5*x^2*deriv(F)/F); return(polcoeff(F,n,x))}
%Y Cf. A008546, A112941 (log derivative); A112934, A112935, A112936, A112937, A112938, A112939, A112942, A112943.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Oct 09 2005