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A316784
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Number of orderless identity tree-factorizations of n.
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2
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1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 4, 1, 2, 2, 3, 1, 4, 1, 4, 2, 2, 1, 10, 1, 2, 2, 4, 1, 8, 1, 6, 2, 2, 2, 13, 1, 2, 2, 10, 1, 8, 1, 4, 4, 2, 1, 26, 1, 4, 2, 4, 1, 10, 2, 10, 2, 2, 1, 28, 1, 2, 4, 13, 2, 8, 1, 4, 2, 8, 1, 46, 1, 2, 4, 4, 2, 8, 1, 26, 3, 2, 1
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OFFSET
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1,6
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COMMENTS
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A factorization of n is a finite nonempty multiset of positive integers greater than 1 with product n. An orderless identity tree-factorization of n is either (case 1) the number n itself or (case 2) a finite set of two or more distinct orderless identity tree-factorizations, one of each factor in a factorization of n.
a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018
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LINKS
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FORMULA
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EXAMPLE
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The a(24)=10 orderless identity tree-factorizations:
24
(4*6)
(3*8)
(2*12)
(2*3*4)
(4*(2*3))
(3*(2*4))
(2*(2*6))
(2*(3*4))
(2*(2*(2*3)))
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MATHEMATICA
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postfacs[n_]:=If[n<=1, {{}}, Join@@Table[Map[Prepend[#, d]&, Select[postfacs[n/d], Min@@#>=d&]], {d, Rest[Divisors[n]]}]];
oltsfacs[n_]:=If[n<=1, {{}}, Prepend[Select[Union@@Function[q, Sort/@Tuples[oltsfacs/@q]]/@DeleteCases[postfacs[n], {n}], UnsameQ@@#&], n]];
Table[Length[oltsfacs[n]], {n, 100}]
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PROG
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(PARI) seq(n)={my(v=vector(n), w=vector(n)); w[1]=v[1]=1; for(k=2, n, w[k]=v[k]+1; forstep(j=n\k*k, k, -k, my(i=j, e=0); while(i%k==0, i/=k; e++; v[j] += binomial(w[k], e)*v[i]))); w} \\ Andrew Howroyd, Nov 18 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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