A291899 -id:A291899 - cp's OEIS frontend

A137927 a(n) = the largest divisor of A000005(n) that is coprime to n. (A000005(n) = the number of positive divisors of n.).

1, 1, 2, 3, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 4, 5, 2, 1, 2, 3, 4, 1, 2, 1, 3, 1, 4, 3, 2, 1, 2, 3, 4, 1, 4, 1, 2, 1, 4, 1, 2, 1, 2, 3, 2, 1, 2, 5, 3, 3, 4, 3, 2, 1, 4, 1, 4, 1, 2, 1, 2, 1, 2, 7, 4, 1, 2, 3, 4, 1, 2, 1, 2, 1, 2, 3, 4, 1, 2, 1

Offset: 1

Leroy Quet, Feb 23 2008

nonn

Apparently also the denominator of A007955(n)/A000005(n). See A291186. - Jaroslav Krizek, Sep 05 2017

			20 has 6 positive divisors. The divisors of 6 are 1,2,3,6. The divisors of 6 that are coprime to 20 are 1 and 3. 3 is the largest of these; so a(20) = 3.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A137926, A007955, A120736, A291899.

A137927 := proc(n)
    local a;
    a := 1 ;
    for d in numtheory[divisors](numtheory[tau](n)) do
        if igcd(d,n) = 1 then
            a := max(a,d) ;
        end if:
    end do:
    a ;
end proc:
seq(A137927(n),n=1..100) ; # R. J. Mathar, Sep 22 2017

Table[Select[Divisors[Length[Divisors[n]]], GCD[ #, n] == 1 &][[ -1]], {n, 1, 80}] (* Stefan Steinerberger, Mar 09 2008 *)

a(n) = my(d=divisors(numdiv(n))); forstep(k=#d, 1, -1, if (gcd(d[k], n) == 1, return (d[k]))); \\ Michel Marcus, Sep 22 2017; corrected Jun 13 2022

More terms from Stefan Steinerberger, Mar 09 2008

A318279 a(n) is the least k such that k^(tau(n)-1) >= n^tau(n).

Original entry on oeis.org

4, 9, 8, 25, 11, 49, 16, 27, 22, 121, 20, 169, 34, 37, 32, 289, 33, 361, 37, 58, 62, 529, 38, 125, 78, 81, 55, 841, 49, 961, 64, 106, 111, 115, 57, 1369, 128, 133, 68, 1681, 72, 1849, 94, 97, 165, 2209, 74, 343, 110, 190, 115, 2809, 96, 210, 100, 220, 225

Offset: 2

David A. Corneth, Oct 09 2018

For prime p and m > 0, a(p^m) = p^(m+1). - Muniru A Asiru, Nov 23 2018

			As tau(4) = 3, we look for the least k such that k^(3-1) >= 4^3, for which we find k = 8. Therefore, a(4) = 8.

David A. Corneth, Table of n, a(n) for n = 2..10001 (first 499 terms by Muniru A Asiru)

Cf. A000005, A291899.

List(List([2..57],n->Filtered([2..3000],k->k^(Tau(n)-1) >= n^Tau(n))),i->i[1]); # Muniru A Asiru, Oct 09 2018

Array[Block[{k = 1}, While[k^(#2 - 1) < #1^#2, k++] & @@ {#, DivisorSigma[0, #]}; k] &, 55, 2] (* Michael De Vlieger, Oct 10 2018 *)

a(n) = my(nd = numdiv(n)); res = ceil(n ^ (nd / (nd - 1))); while(res^(nd-1) >= n^nd, res--); res+1

a(n) = ceiling(n^(1 + 1/(tau(n)-1))). - Jon E. Schoenfield, Nov 22 2018

Correct value a(27) = 81 inserted by Muniru A Asiru, Nov 22 2018

cp's OEIS Frontend

A137927 a(n) = the largest divisor of A000005(n) that is coprime to n. (A000005(n) = the number of positive divisors of n.).

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