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%I #14 Nov 19 2018 03:11:43
%S 1,1,0,1,1,1,0,0,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,0,1,0,1,1,4,1,0,1,1,1,
%T 3,1,1,1,1,1,4,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,9,1,1,1,0,1,4,1,1,1,
%U 4,1,6,1,1,1,1,1,4,1,1,0,1,1,9,1,1,1,1,1,9,1
%N Number of complete strict tree-factorizations of n > 1.
%C A strict tree-factorization (see A295279 for definition) is complete if its leaves are all prime numbers.
%C From _Andrew Howroyd_, Nov 18 2018: (Start)
%C a(n) depends only on the prime signature of n.
%C This sequence is very similar but not identical to the number of complete orderless identity tree-factorizations of n. The first difference is at n=900 (square of three primes). Here a(n) = 191 whereas the other sequence would have 197. (End)
%H Andrew Howroyd, <a href="/A295281/b295281.txt">Table of n, a(n) for n = 2..10000</a>
%F a(product of n distinct primes) = A000311(n).
%F Positions of zeros are proper prime powers A025475. Positions of nonzero entries are A085971.
%e The a(72) = 6 complete strict tree-factorizations are: 2*3*(2*(2*3)), 2*(2*3*(2*3)), 2*(2*(3*(2*3))), 2*(3*(2*(2*3))), 3*(2*(2*(2*3))), (2*3)*(2*(2*3)).
%t postfacs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[postfacs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
%t sftc[n_]:=Prepend[Join@@Function[fac,Tuples[sftc/@fac]]/@Select[postfacs[n],And[Length[#]>1,UnsameQ@@#]&],n];
%t Table[Length[Select[sftc[n],FreeQ[#,_Integer?(!PrimeQ[#]&)]&]],{n,2,100}]
%o (PARI) seq(n)={my(v=vector(n), w=vector(n)); v[1]=1; for(k=2, n, w[k]=v[k]+isprime(k); forstep(j=n\k*k, k, -k, v[j]+=w[k]*v[j/k])); w[2..n]} \\ _Andrew Howroyd_, Nov 18 2018
%Y Cf. A000311, A025475, A085971, A273873, A281113 A281118, A281119, A292505, A295279.
%K nonn
%O 2,29
%A _Gus Wiseman_, Nov 19 2017
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