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a(0) = 1; a(n) = (1/2) * Sum_{k=1..n} (3^k-1) * a(k-1) * a(n-k).
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%I #5 Sep 11 2024 00:44:54

%S 1,1,5,74,3119,384099,140605620,153966205482,505318125737963,

%T 4973847408741044519,146857822147450491641165,

%U 13007931631590001724722114996,3456493610037973055076316970551876,2755388815749774181719259556096183210356,6589473777446361501832833785593366614276353520

%N a(0) = 1; a(n) = (1/2) * Sum_{k=1..n} (3^k-1) * a(k-1) * a(n-k).

%F G.f. A(x) satisfies: A(x) = 2 / (2 + x * A(x) - 3 * x * A(3*x)).

%t a[0] = 1; a[n_] := a[n] = (1/2) Sum[(3^k - 1) a[k - 1] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 14}]

%t nmax = 14; A[_] = 0; Do[A[x_] = 2/(2 + x A[x] - 3 x A[3 x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%Y Cf. A015084, A337555, A376111, A376113.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Sep 10 2024