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a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} (-2)^k * a(k) * a(n-2*k-1).
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%I #6 Mar 01 2022 07:28:18

%S 1,1,1,-1,-3,-1,5,23,29,-33,-139,-217,13,943,1765,-1545,-8963,-11265,

%T 6229,73671,126701,-65713,-567611,-793449,415197,4231583,7471669,

%U -4933529,-37928499,-52823313,28920485,346647351,610476733,-316142785,-2913394667,-4922323705

%N a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} (-2)^k * a(k) * a(n-2*k-1).

%F G.f. A(x) satisfies: A(x) = 1 / (1 - x * A(-2*x^2)).

%t a[0] = 1; a[n_] := a[n] = Sum[(-2)^k a[k] a[n - 2 k - 1], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 35}]

%t nmax = 35; A[_] = 0; Do[A[x_] = 1/(1 - x A[-2 x^2]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%Y Cf. A015097, A101912, A352010, A352011.

%K sign

%O 0,5

%A _Ilya Gutkovskiy_, Feb 28 2022