login
a(n) = n / gcd(n, A163511(n)).
8

%I #15 Jul 27 2023 08:18:23

%S 0,1,1,1,1,5,1,7,1,1,5,11,1,13,7,15,1,17,1,19,5,7,11,23,1,5,13,27,7,

%T 29,15,31,1,11,17,7,1,37,19,39,5,41,7,43,11,15,23,47,1,49,5,51,13,53,

%U 27,5,7,19,29,59,15,61,31,63,1,65,11,67,17,23,7,71,1,73,37,15,19,11,39,79,5,3,41,83,7,17,43,87

%N a(n) = n / gcd(n, A163511(n)).

%C Numerator of n / A163511(n).

%H Antti Karttunen, <a href="/A364491/b364491.txt">Table of n, a(n) for n = 0..16383</a>

%F a(n) = n / A364255(n) = n / gcd(n, A163511(n)).

%o (PARI)

%o A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t };

%o A054429(n) = ((3<<#binary(n\2))-n-1); \\ From A054429

%o A163511(n) = if(!n,1,A005940(1+A054429(n)))

%o A364491(n) = (n/gcd(n, A163511(n)));

%o (Python)

%o from math import gcd

%o from sympy import nextprime

%o def A364491(n):

%o c, p, k = 1, 1, n

%o while k:

%o c *= (p:=nextprime(p))**(s:=(~k&k-1).bit_length())

%o k >>= s+1

%o return n//gcd(c*p,n) # _Chai Wah Wu_, Jul 26 2023

%Y Cf. A163511, A364255, A364492 (denominators), A364493, A364494 (positions of 1's).

%K nonn,frac

%O 0,6

%A _Antti Karttunen_, Jul 26 2023