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Sum_{i=0..n} Sum_{j=0..i} a(j) * a(i-j) = (-7)^n.
3

%I #14 Feb 04 2023 10:18:31

%S 1,-4,20,-116,708,-4452,28532,-185300,1215268,-8030404,53381844,

%T -356577588,2391430020,-16092704292,108605848116,-734783381652,

%U 4982063186916,-33844621986180,230306722637204,-1569571734301172,10711405584991300,-73188920628617956,500643475619050740

%N Sum_{i=0..n} Sum_{j=0..i} a(j) * a(i-j) = (-7)^n.

%H Seiichi Manyama, <a href="/A085456/b085456.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: A(x)=Sqrt((1-x)/(1+7x)).

%F From _Seiichi Manyama_, Feb 03 2023: (Start)

%F a(n) = Sum_{k=0..n} (-2)^k * binomial(n-1,n-k) * binomial(2*k,k).

%F n*a(n) = -2*(3*n-1)*a(n-1) + 7*(n-2)*a(n-2). (End)

%t CoefficientList[Series[Sqrt[(1-x)/(1+7 x)], {x, 0, 30}], x]

%o (PARI) a(n) = sum(k=0, n, (-2)^k*binomial(n-1, n-k)*binomial(2*k, k)); \\ _Seiichi Manyama_, Feb 03 2023

%Y Cf. A085455, A085457, A085458.

%K easy,sign

%O 0,2

%A Mario Catalani (mario.catalani(AT)unito.it), Jul 01 2003