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A338681
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The number of factorizations of an n-element set. (Defined below in Comments.)
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0
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1, 1, 1, 4, 1, 61, 1, 1681, 5041, 15121, 1, 13638241, 1, 8648641, 1816214401, 181880899201, 1, 45951781075201, 1, 3379365788198401, 1689515283456001, 14079294028801, 1, 4454857103544668620801, 538583682060103680001, 32382376266240001
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OFFSET
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1,4
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COMMENTS
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A factorization of a set S is a set B of nontrivial partitions of S such that for each way of choosing one part from each partition in B, there exists a unique element of S in the intersection of the chosen parts.
A factorization of a set can be thought of as a multiplicative analog of a set partition, so this sequence can be thought of as a multiplicative analog of the Bell numbers (A000110).
a(p)=1 for p prime.
For all positive integers k, a(n) = 1 (mod k) for all sufficiently large n.
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LINKS
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FORMULA
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Let T_n be the set of all lists a_1b_1...a_kb_k of positive integers, where k >= 0, n = a_1^b_1*...*a_k^b_k, and for all j, a_j >= 2 and a_j > a_{j+1}. (Note that T_n can be thought of as the set of multiplicative partitions of n, so |T_n| = A001055(n).) Then A(n) equals the sum over all a_1b_1...a_kb_k in T_n of n!/Product_{j=1..k} ((a_j!^b_j)*b^j!).
a(n) = n!*R(n,0) where R(1,k) = 1/k! and R(n,k) = Sum_{d|n, d>1} R(n/d, k+1)/d!. - Andrew Howroyd, May 11 2021
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EXAMPLE
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For n = 4, the four factorizations of {0,1,2,3} are {{{0},{1},{2},{3}}}, {{{0,1},{2,3}},{{0,2},{1,3}}}, {{{0,1},{2,3}},{{0,3},{1,2}}}, and {{{0,2},{1,3}},{{0,3},{1,2}}}.
a(6) = 61 because there is 1 solution {{{0},{1},{2},{3},{4],{5}}} and 60 = 10 * 6 of the form {{{a,b,c}, {d,e,f}}, {{a,d},{b,e},{c,f}}}.
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MATHEMATICA
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R[n_, k_] := R[n, k] = If[n == 1, 1/k!, Sum[If[d > 1, R[n/d, k+1]/d!, 0], {d, Divisors[n]}]];
a[n_] := n! R[n, 0];
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PROG
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(PARI)
R(n, k)={if(n==1, 1/k!, sumdiv(n, d, if(d>1, self()(n/d, k+1)/d! )))}
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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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