A052128 a(1) = 1; for n > 1, a(n) is the largest divisor of n that is coprime to a larger divisor of n.

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

Offset: 1

Author

James Sellers, Jan 21 2000

Keywords

• nonn

Comments

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

From Jianing Song, Sep 28 2022: (Start)

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)

Examples

			a(6) = 6 / 3^1 = 2.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A054372, A076388, A354933.

Programs

Mathematica

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 *)

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

Formula

a(n) = n / A354933(n) = A354933(n) - A076388(n). - Antti Karttunen, Jun 16 2022

Extensions

Terms corrected by Charles R Greathouse IV, Aug 22 2013

Definition rewritten by Jianing Song, Sep 28 2022