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A052128
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a(1) = 1; for n > 1, a(n) is the largest divisor of n that is coprime to a larger divisor of n.
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6
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1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 3, 1, 1, 2, 1, 4, 3, 2, 1, 3, 1, 2, 1, 4, 1, 5, 1, 1, 3, 2, 5, 4, 1, 2, 3, 5, 1, 6, 1, 4, 5, 2, 1, 3, 1, 2, 3, 4, 1, 2, 5, 7, 3, 2, 1, 5, 1, 2, 7, 1, 5, 6, 1, 4, 3, 7, 1, 8, 1, 2, 3, 4, 7, 6, 1, 5, 1, 2, 1, 7, 5, 2, 3, 8, 1, 9, 7, 4, 3, 2, 5, 3, 1, 2, 9, 4, 1, 6
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OFFSET
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1,6
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COMMENTS
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Least k > 0 such that the resultant of the k-th cyclotomic polynomial and the n-th cyclotomic polynomial is not 1. - Benoit Cloitre, Oct 13 2002
a(n) is the largest divisor d of n such that d <= sqrt(n) and that gcd(d,n/d) = 1.
Proof: write n = Product_{1<=i<=r} (p_i)^(e_i), let d be the largest divisor of n such that d <= sqrt(n) and that gcd(d,n/d) = 1. Obviously we have a(n) >= d. Suppose that a(n) = Product_{1<=i<=s} (p_i)^(m_i) for s <= r, 1 <= m_i <= e_i, then the larger divisor to which a(n) is coprime is a divisor of Product_{s+1<=i<=r} (p_i)^(e_i), so by definition we have a(n) <= min{Product_{1<=i<=s} (p_i)^(e_i), Product_{s+1<=i<=r} (p_i)^(e_i)} <= d. Thus a(n) = d. (End)
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LINKS
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FORMULA
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EXAMPLE
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a(6) = 6 / 3^1 = 2.
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MATHEMATICA
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Table[best = 1; d = Divisors[n]; While[Length[d] > 1, e = d[[1]]; d = Rest[d]; If[Min[GCD[e, d]] == 1, best = e]]; best, {n, 102}] (* T. D. Noe, Aug 23 2013 *)
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PROG
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(PARI) a(n) = my(i, j, d = divisors(n)); forstep (i = #d-1, 1, -1, for (j = i+1, #d, if (gcd(d[i], d[j]) == 1, return (d[i])))); 1 \\ Michel Marcus, Aug 22 2013
(PARI) a(n)=my(f=factor(n), v=[1]); for(i=1, #f~, v=concat(v, f[i, 1]^f[i, 2] *v)); v=vecsort(v); forstep(i=#v\2, 2, -1, for(j=i+1, #v-1, if(gcd(v[i], v[j])==1, return(v[i])))); 1 \\ Charles R Greathouse IV, Aug 22 2013
(PARI) A052128(n) = fordiv(n, d, if((d>=(n/d)) && 1==gcd(d, n/d), return(n/d))); \\ Antti Karttunen, Jun 16 2022
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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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