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G.f. A(x) satisfies: A(x) = 1 / (1 + x - 2 * x * A(2*x)).
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%I #12 Jul 15 2024 12:12:14

%S 1,1,5,49,893,30649,2030213,264198625,68180168717,35046644401609,

%T 35958357173552597,73714882938928013809,302083844634245306686685,

%U 2475275541582550287356775001,40559867144321249927245807932197,1329146863668196853655964629931680001

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

%H Michael De Vlieger, <a href="/A348901/b348901.txt">Table of n, a(n) for n = 0..81</a>

%H Johann Cigler, <a href="https://arxiv.org/abs/2407.05768">Hankel determinants of backward shifts of powers of q</a>, arXiv:2407.05768 [math.CO], 2024. See p. 6.

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

%F a(n) ~ 2^(n*(n+1)/2). - _Vaclav Kotesovec_, Nov 03 2021

%t nmax = 15; A[_] = 0; Do[A[x_] = 1/(1 + x - 2 x A[2 x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t a[0] = 1; a[n_] := a[n] = -a[n - 1] + Sum[2^(k + 1) a[k] a[n - k - 1], {k, 0, n - 1}]; Table[a[n], {n, 0, 15}]

%Y Cf. A001003, A015083, A348188, A348902.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Nov 03 2021