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A316789
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Number of same-tree-factorizations of n.
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2
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1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 14, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1
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OFFSET
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1,4
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COMMENTS
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A constant factorization of n is a finite nonempty constant multiset of positive integers greater than 1 with product n. Constant factorizations correspond to perfect divisors (A089723). A same-tree-factorization of n is either (case 1) the number n itself or (case 2) a finite sequence of two or more same-tree-factorizations, one of each factor in a constant 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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a(n) = 1 + Sum_{n = x^y, y > 1} a(x)^y.
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EXAMPLE
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The a(64) = 14 same-tree-factorizations:
64
(8*8)
(4*4*4)
(8*(2*2*2))
((2*2*2)*8)
(4*4*(2*2))
(4*(2*2)*4)
((2*2)*4*4)
(2*2*2*2*2*2)
(4*(2*2)*(2*2))
((2*2)*4*(2*2))
((2*2)*(2*2)*4)
((2*2*2)*(2*2*2))
((2*2)*(2*2)*(2*2))
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MATHEMATICA
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a[n_]:=1+Sum[a[n^(1/d)]^d, {d, Rest[Divisors[GCD@@FactorInteger[n][[All, 2]]]]}]
Array[a, 100]
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PROG
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(PARI) a(n)={my(z, e=ispower(n, , &z)); 1 + if(e, sumdiv(e, d, if(d>1, a(z^(e/d))^d)))} \\ Andrew Howroyd, Nov 18 2018
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CROSSREFS
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Cf. A001055, A001597, A001678, A003238, A007916, A052409, A052410, A067824, A089723, A281118, A281145, A294336, A316790.
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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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