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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)=1, k=-2 and l=0.
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%I #11 Jan 03 2024 16:22:48

%S 1,1,-2,-9,-30,-84,-204,-389,-326,1780,13156,57452,197552,551846,

%T 1138832,752911,-8109806,-57353648,-255573404,-898715548,-2539157248,

%U -5106161134,-1629827488,50275158584,330772150256,1453122571658

%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)=1, k=-2 and l=0.

%H Robert Israel, <a href="/A177111/b177111.txt">Table of n, a(n) for n = 0..1767</a>

%F G.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=-2, l=0).

%F Conjecture: +(n+1)*a(n) +(-7*n+2)*a(n-1) +3*(5*n-7)*a(n-2) +3*(n-8)*a(n-3) +2*(-10*n+41)*a(n-4) +8*(n-5)*a(n-5)=0. - _R. J. Mathar_, Mar 02 2016

%F Conjectured recurrence follows from the differential equation (8*z^6 - 20*z^5 + 3*z^4 + 15*z^3 - 7*z^2 + z) * f'(z) + (2*z^4 - 15*z^3 + 9*z^2 - 5*z + 1) * f(z) + 4*z^4 - 11*z^3 + 9*z^2 + 3*z - 1 = 0 satisfied by the g.f. - _Robert Israel_, Jan 03 2024

%e a(2)=2*1*1-4=-2. a(3)=2*1*(-2)-4+1^2-2=-9. a(4)=2*1*(-9)-4+2*1*(-2)-4=-30.

%p l:=0: : k := -2 : m:=1:d(0):=1:d(1):=m: for n from 1 to 30 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, 30); seq(d(n), n=0..30);

%Y Cf. A177110.

%K easy,sign

%O 0,3

%A _Richard Choulet_, May 03 2010