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Search: id:A104725
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| A104725 |
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Number of complementing systems of subsets of {0, 1, ..., n-1}. |
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2
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| 0, 1, 1, 1, 2, 1, 3, 1, 5, 2, 3, 1, 11, 1, 3, 3, 15, 1, 11, 1, 11, 3, 3, 1, 45, 2, 3, 5, 11, 1, 19, 1, 52, 3, 3, 3, 62, 1, 3, 3, 45, 1, 19, 1, 11, 11, 3, 1, 200, 2, 11, 3, 11, 1, 45, 3, 45, 3, 3, 1, 113, 1, 3, 11, 203, 3, 19, 1, 11, 3, 19, 1, 355, 1, 3, 11, 11, 3, 19, 1, 200, 15, 3, 1, 113, 3
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OFFSET
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0,5
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COMMENT
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Number of collections {S_1, S_2, ..., S_k} of subsets of {0, 1, ..., n-1}, each subset containing 0, such that every element x of {0,1, ..., n-1} can be uniquely expressed as x=x_1+x_2+ ...+ x_k with x_i in S_i for all i=1,2,..,k.
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REFERENCES
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C. T. Long, Addition Theorems for sets of Integers, Pacific J. Math. 23 (1967), 107-112.
A. O. Munagi, k-Complementing Subsets of Nonnegative Integers, IJMMS 2005:2, (2005), 215-224.
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LINKS
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N. J. A. Sloane, Table of n, a(n) for n = 0..10000
A. O. Munagi, Notes on A104725
A. O. Munagi, k-Complementing Subsets of Nonnegative Integers, IJMMS 2005:2 (2005), 215-224.
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FORMULA
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a(0)=0, a(1)=1, a(n)=Sum(ordfac(n,k)*Bell(k-1),k=1..Omega(n)), n>1, where ordfac(n,k)=number of ordered factorizations of n into k factors.
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EXAMPLE
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a(6) = 3: {{0,1,2,3,4,5}}, {{0,1,2},{0,3}} and {{0,1},{0,2,4}}.
Thus since {{0,1,2},{0,3}} is a complementing system of subsets of {0,1,2,3,4,5} we have 0=0+0, 1=1+0, 2=2+0, 3=0+3, 4=1+3, 5=2+3.
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MAPLE
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a:=proc(n::integer) local u, r, i, j, k; if n<1 then return 0; elif n=1 then return 1; end if; u:=map(x->x[2], ifactors(n)[2]); r:=add(u[i], i=1..nops(u)); add(add((-1)^i*binomial(k, i)*product(binomial(u[j]+k-i-1, u[j]), j=1..nops(u)), i=0..k-1)*bell(k-1), k=1..r); end proc: seq(a(n), n=0..90);
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CROSSREFS
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Cf. A074206, A002033. a(n)= A074206(n) if A001222(n)=1, 2.
Sequence in context: A129982 A052552 A085053 this_sequence A011129 A065370 A102614
Adjacent sequences: A104722 A104723 A104724 this_sequence A104726 A104727 A104728
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KEYWORD
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nonn,nice,core
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AUTHOR
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A. O. Munagi (amunagi(AT)yahoo.com), Mar 20, 2005; Dec 20, 2006
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