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Number of ordered factorizations of n into Lucas numbers (beginning at 2) > 1.
0

%I #5 May 17 2019 22:07:48

%S 1,1,1,2,0,2,1,3,1,0,1,5,0,2,0,5,0,4,0,0,2,2,0,10,0,0,1,5,1,0,0,8,2,0,

%T 0,11,0,0,0,0,0,6,0,5,0,0,1,20,1,0,0,0,0,6,0,10,0,2,0,0,0,0,3,13,0,6,

%U 0,0,0,0,0,27,0,0,0,1,2,0,0,0,1,0,0,18,0,0,2,10

%N Number of ordered factorizations of n into Lucas numbers (beginning at 2) > 1.

%F G.f. A(x) satisfies: A(x) = x + A(x^2) + Sum_{k>=2} A(x^Lucas(k)).

%t terms = 88; A[_] = 0; Do[A[x_] = x + A[x^2] + Sum[A[x^LucasL[k]], {k, 2, 25}] + O[x]^(terms + 1) // Normal, terms + 1]; Rest[CoefficientList[A[x], x]]

%t f[n_] := f[n] = SeriesCoefficient[x^2 + Sum[x^LucasL[k], {k, 2, 25}], {x, 0, n}]; a[n_] := If[n == 1, n, Sum[If[d < n, f[n/d] a[d], 0], {d, Divisors[n]}]]; Table[a[n], {n, 1, 88}]

%Y Cf. A000032, A102460, A200381 (positions of nonzero terms), A200995 (positions of zeros), A308062.

%K nonn

%O 1,4

%A _Ilya Gutkovskiy_, May 17 2019