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Sequence defined by the recurrence formula a(n+1) = Sum_{n>=1}(a(p)*a(n-p) + k, p=0..n) + j, with a(0) = 1, a(1) = 0, k = 0 and j = 1.
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%I #21 Jan 26 2020 16:33:48

%S 1,0,1,3,7,16,39,102,279,782,2227,6427,18769,55376,164801,494071,

%T 1490663,4522690,13790171,42234621,129866725,400765128,1240796725,

%U 3853055776,11997619209,37451945874,117181432493,367428949069

%N Sequence defined by the recurrence formula a(n+1) = Sum_{n>=1}(a(p)*a(n-p) + k, p=0..n) + j, with a(0) = 1, a(1) = 0, k = 0 and j = 1.

%C The link contains a list of all 85 related sequences and their parameters a(1)=m, k and j, together with a proof of the recurrence given by Richard J. Mathar. - _Georg Fischer_, Jan 26 2020

%H Georg Fischer, <a href="/A176604/b176604.txt">Table of n, a(n) for n = 0..200</a>

%H Georg Fischer, <a href="/A176604/a176604_2.txt">Derivation of the D-finite recurrence equation</a> for A176604 and related sequences.

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

%F (n+1)*a(n) +2*(-3*n+1)*a(n-1) +(13*n-21)*a(n-2) +16*(-n+3)*a(n-3) +8*(n-4)*a(n-4)=0. - _R. J. Mathar_, Feb 19 2016

%e a(2) = (a(0)*a(1)+0)+(a(1)*a(0)+0)+1 = 1.

%e a(3) = (a(0)*a(2)+0)+(a(1)^2+0)+(a(2)*a(0)+0)+1 = 3.

%e a(4) = 2*a(0)*a(3)+2*a(1)*a(2)+1 = 7.

%t (* Applicable for all 85 related sequences *)

%t m:=0; k:=0; j:=1; CoefficientList[Series[(1-Sqrt[1-4*z*(1-z+z*m+(k+j)*z^2/(

%t 1-z)+k*z^2/(1-z)^2)])/(2*z),{z,0,20}],z]

%t (* or *)

%t m:=0; k:=0; j:=1;

%t RecurrenceTable[{a[0]==1, a[1]==m, a[2]==1, a[3]==3, a[4]==7,

%t + (+20+20*k+20*j-20*m+(- 4- 4*k-4*j+ 4*m)*n)*a[n-5]

%t - (+62+48*k+34*j-48*m+(-16-12*k-8*j+12*m)*n)*a[n-4]

%t + (+68+28*k+14*j-36*m+(-25- 8*k-4*j+12*m)*n)*a[n-3]

%t - (+29 - 8*m+(-19 + 4*m)*n)*a[n-2]

%t - ( -2 +( 7 )*n)*a[n-1]

%t + ( +1 +( 1 )*n)*a[n ]

%t == 0},a,{n,0,20}]== 0},a,{n,0,20}] (* _Georg Fischer_, Jan 26 2020 - cf. link *)

%K easy,nonn

%O 0,4

%A _Richard Choulet_, Apr 21 2010