A125168 a(n) = gcd(n, A032741(n)) where A032741(n) is the number of proper divisors of n.

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

Offset: 1

Author

Mitch Cervinka (puritan(AT)toast.net), Jan 12 2007

Keywords

• easy
• nonn

Comments

First occurrence of k: 1, 4, 6, 16, 20, 3240000, 42, 256, 162, 18662400, 132, 5308416, 832, 784, 120, 65536, 612, 2985984, 912, 1600, 9240, 98010000, 1380, 1296, 100800, ..., (10^7). - Robert G. Wilson v, Jan 23 2007

Do all values appear? - Robert G. Wilson v, Jan 23 2007

From Bernard Schott, Oct 19 2019: (Start)

a(n) = 1 if n = p^k, p prime, k >= 0 and k <> p or,

n = p*q, p 3 or

n = p*q*r, p 7 or,

n = p^2*q, p 5 or

n = p^3*q, p 7.

a(n) = 2 if n = 2^2 or n = 2^(2*p), p prime <> 2,

a(n) = 3 if n = 3*p, p prime <> 3 or n = 3^3,

a(n) = 4 if n = 4*p^2, p prime,

a(n) = 5 if n = 5*p^2, p prime <> 5, or n = 25*p, p prime <> 5, or n = 5^5,

a(n) = 7 if n = 7*p*q with p 7 or n = 7*p^3, p prime <> 7, or n = 7^7,

a(n) = p if n = p^p, p prime. (End)

Examples

			a(6)=3 because 6 has 3 proper divisors {1,2,3} and gcd(6,3) is 3.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A032741, A009191, A062968.

Programs

Mathematica

f[n_] := GCD[n, DivisorSigma[0, n] - 1]; Array[f, 105] (* Robert G. Wilson v *)

PARI

A125168(n) = gcd(n,numdiv(n)-1); \\ Antti Karttunen, Sep 25 2018

Formula

a(n) = gcd(n, A032741(n)) = gcd(n, A062968(n)).

Extensions

More terms from Robert G. Wilson v, Jan 23 2007