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a(1) = a(2) = 1; a(n+2) = Sum_{d|n, d < n} a(d).
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%I #6 Jun 09 2021 11:14:26

%S 1,1,0,1,1,2,1,2,1,3,1,3,1,5,1,3,2,5,1,5,1,7,2,3,1,10,2,3,2,9,1,9,1,8,

%T 2,4,3,14,1,3,2,14,1,11,1,11,4,4,1,16,2,7,3,14,1,12,3,14,2,4,1,27,1,3,

%U 4,17,3,13,1,13,3,14,1,23,1,5,4,18,3,16,1,20,4,4,1,32,4,3,3,24,1,25,3,16,2

%N a(1) = a(2) = 1; a(n+2) = Sum_{d|n, d < n} a(d).

%F G.f. A(x) satisfies: A(x) = x + x^2 * (1 + A(x^2) + A(x^3) + A(x^4) + ...).

%t a[1] = a[2] = 1; a[n_] := a[n] = Sum[If[d < n - 2, a[d], 0], {d, Divisors[n - 2]}]; Table[a[n], {n, 1, 95}]

%t nmax = 95; A[_] = 0; Do[A[x_] = x + x^2 (1 + Sum[A[x^k], {k, 2, nmax}]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] // Rest

%Y Cf. A007439, A074206, A167865, A345138, A345141.

%K nonn

%O 1,6

%A _Ilya Gutkovskiy_, Jun 09 2021