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Sequence defined by the recurrence formula a(n+1)=sum(a(p)*a(n-p)+k,p=0..n)+l for n>=1, with here a(0)=1, a(1)=8, k=0 and l=1.
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%I #5 Jun 14 2016 12:40:09

%S 1,8,17,99,471,2816,16535,103942,661447,4327566,28698915,193214427,

%T 1314753729,9035450112,62597834193,436806174807,3066961374135,

%U 21653065678706,153619938907211,1094646596551549,7830810922793173

%N Sequence defined by the recurrence formula a(n+1)=sum(a(p)*a(n-p)+k,p=0..n)+l for n>=1, with here a(0)=1, a(1)=8, k=0 and l=1.

%F G.f f: f(z)=(1-sqrt(1-4*z*(a(0)-z*a(0)^2+z*a(1)+(k+l)*z^2/(1-z)+k*z^2/(1-z)^2)))/(2*z) (k=0, l=1).

%F Conjecture: (n+1)*a(n) +2*(-3*n+1)*a(n-1) +(-19*n+43)*a(n-2) +48*(n-3)*a(n-3) +24*(-n+4)*a(n-4)=0. - _R. J. Mathar_, Jun 14 2016

%e a(2)=2*1*8+1=17. a(3)=2*1*17+64+1=99.

%p l:=1: : k := 0 : m :=8: d(0):=1:d(1):=m: for n from 1 to 28 do d(n+1):=sum(d(p)*d(n-p)+k, p=0..n)+l:od :

%p taylor((1-sqrt(1-4*z*(d(0)-z*d(0)^2+z*m+(k+l)*z^2/(1-z)+k*z^2/(1-z)^2)))/(2*z), z=0, 31); seq(d(n), n=0..29);

%Y Cf. A177128.

%K easy,nonn

%O 0,2

%A _Richard Choulet_, May 03 2010