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A085455 Sum_{i=0..n} Sum_{j=0..i} a(j) * a(i-j) = (-3)^n. 5

%I #14 Feb 04 2023 10:18:24

%S 1,-2,4,-10,26,-70,192,-534,1500,-4246,12092,-34606,99442,-286730,

%T 829168,-2403834,6984234,-20331558,59287740,-173149662,506376222,

%U -1482730098,4346486256,-12754363650,37461564504,-110125172682,323990062452,-953883382354,2810310510110,-8284915984726

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

%H Seiichi Manyama, <a href="/A085455/b085455.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: A(x)=Sqrt((1-x)/(1+3x)).

%F G.f.: G(0), where G(k)= 1 + 4*x*(4*k+1)/( (x-1)*(4*k+2) - x*(x-1)*(4*k+2)*(4*k+3)/(x*(4*k+3) + (x-1)*(k+1)/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 26 2013

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

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

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

%t CoefficientList[Series[Sqrt[(1-x)/(1+3x)], {x, 0, 30}], x]

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

%Y Absolute values are in A025565.

%K easy,sign

%O 0,2

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

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Last modified May 11 11:07 EDT 2024. Contains 372409 sequences. (Running on oeis4.)