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a(n) is the smallest positive integer that is coprime to n and has n divisors.
2

%I #21 Mar 27 2022 13:37:20

%S 1,3,4,15,16,175,64,105,100,567,1024,1925,4096,3645,784,945,65536,

%T 13475,262144,6237,1600,295245,4194304,25025,1296,2657205,4900,40095,

%U 268435456,3776773,1073741824,10395,25600,215233605,5184,175175,68719476736,1937102445,102400

%N a(n) is the smallest positive integer that is coprime to n and has n divisors.

%C Is this the same as the least index m where A090387(m) = n? - _Michel Marcus_, Mar 25 2022

%C For p prime, a(p) = 2^(p-1) for p > 2, a(2*p) = 3^(p-1)*5 for p > 5, a(3*p) = 2^(p-1)*25 for p > 3, a(5*p) = 2^(p-1)*3^4 for p >5, ... . - _Michael S. Branicky_, Mar 26 2022

%H Michael S. Branicky, <a href="/A136641/b136641.txt">Table of n, a(n) for n = 1..65</a>

%e The sequence of positive integers each with 9 divisors starts: 36, 100, 196, 225, 256, ... Now 36 is not coprime to 9. But 100, the next bigger value with 9 divisors, is. So a(9) = 100.

%o (PARI) a(n) = my(k=1); while ((gcd(n,k) != 1) || (numdiv(k) != n), k++); k; \\ _Michel Marcus_, Mar 25 2022

%o (Python)

%o from math import gcd

%o from sympy import divisor_count

%o def a(n):

%o k = 1

%o while gcd(n, k) != 1 or divisor_count(k) != n: k += 1

%o return k

%o print([a(n) for n in range(1, 19)]) # _Michael S. Branicky_, Mar 25 2022

%Y Cf. A000005, A073904, A090387.

%K nonn

%O 1,2

%A _Leroy Quet_, Apr 14 2008

%E a(11)-a(36) from _Sean A. Irvine_, May 03 2010

%E a(37) and beyond from _Michael S. Branicky_, Mar 26 2022