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A069023
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Define a subset of divisors of n to be a dedicated subset if the product of any two members is also a divisor of n. 1 is not allowed as a member as it gives trivially 1*d = d a divisor. a(n) is the number of dedicated subsets of divisors of n with at least two members.
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1
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0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 3, 0, 1, 1, 2, 0, 3, 0, 3, 1, 1, 0, 9, 0, 1, 1, 3, 0, 7, 0, 5, 1, 1, 1, 9, 0, 1, 1, 9, 0, 7, 0, 3, 3, 1, 0, 17, 0, 3, 1, 3, 0, 9, 1, 9, 1, 1, 0, 20, 0, 1, 3, 8, 1, 7, 0, 3, 1, 7, 0, 28, 0, 1, 3, 3, 1, 7, 0, 17, 2, 1, 0, 20, 1, 1, 1, 9, 0, 20, 1, 3, 1, 1, 1, 35, 0, 3, 3, 9, 0, 7
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OFFSET
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1,12
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COMMENTS
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a(n) is determined by the prime signature of n.
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LINKS
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FORMULA
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EXAMPLE
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a(12) = 3. The divisors of 12 are 1,2,3,4,6,12. The divisor subsets (2,3),(2,6) and (3,4) are such that their product is also a divisor of 12. a(24) = 9 and the dedicated divisor subsets are (2,3),(2,4),(2,6),(2,12),(3,4),(3,8),(4,6),(2,3,4),(2,4,6).
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PROG
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(PARI)
\\ The following program is very inefficient:
A069023(n) = { if(bigomega(n)<2, return(0)); my(pds=(divisors(n)[2..numdiv(n)]), subsets = select(v -> (length(v)>=2), powerset(pds)), pair_products = apply(ss -> podp(ss), subsets), prodsmodn = apply(pps -> vector(#pps, i, n%pps[i]), pair_products)); length(select(s -> 0==vecsum(s), prodsmodn)); };
powerset(v) = { my(siz=2^length(v), pv=vector(siz)); for(i=0, siz-1, pv[i+1] = choosebybits(v, i)); pv; };
choosebybits(v, m) = { my(s=vector(hammingweight(m)), i=j=1); while(m>0, if(m%2, s[j] = v[i]; j++); i++; m >>= 1); s; };
podp(v) = { my(siz=binomial(length(v), 2), rv=vector(siz), k=0); for(i=1, length(v)-1, for(j=i+1, length(v), k++; rv[k] = v[i]*v[j])); rv; }; \\ podp = product of distinct pairs
(Scheme) ;; See in the links-section.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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