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%I #59 May 29 2021 12:20:57
%S 1,1,1,4,1,61,1,1681,5041,15121,1,13638241,1,8648641,1816214401,
%T 181880899201,1,45951781075201,1,3379365788198401,1689515283456001,
%U 14079294028801,1,4454857103544668620801,538583682060103680001,32382376266240001
%N The number of factorizations of an n-element set. (Defined below in Comments.)
%C 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.
%C 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).
%C a(p)=1 for p prime.
%C For all positive integers k, a(n) = 1 (mod k) for all sufficiently large n.
%F 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!).
%F 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
%e 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}}}.
%e 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}}}.
%t 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]}]];
%t a[n_] := n! R[n, 0];
%t Array[a, 26] (* _Jean-François Alcover_, May 29 2021, after _Andrew Howroyd_ *)
%o (PARI)
%o R(n,k)={if(n==1, 1/k!, sumdiv(n, d, if(d>1, self()(n/d, k+1)/d! )))}
%o a(n)={n!*R(n,0)} \\ _Andrew Howroyd_, May 11 2021
%K nonn
%O 1,4
%A _Scott Garrabrant_, Apr 30 2021
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