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a(n) = n / gcd(n, binomial(n, floor(n/2))).
7

%I #30 Mar 22 2020 09:58:57

%S 1,1,1,2,1,3,1,4,1,5,1,1,1,7,1,8,1,9,1,5,1,11,1,6,1,13,1,7,1,1,1,16,1,

%T 17,1,3,1,19,1,2,1,7,1,11,1,23,1,4,1,25,1,13,1,27,1,1,1,29,1,15,1,31,

%U 1,32,1,11,1,17,1,5,1,18,1,37,1,19,1,39,1,4,1,41,1,1,1,43,1,11,1,1,1,23

%N a(n) = n / gcd(n, binomial(n, floor(n/2))).

%H Hugo Pfoertner, <a href="/A049077/b049077.txt">Table of n, a(n) for n = 1..10000</a>

%F For odd n, a(n) = 1. For even n, a(n) is either n/2 or smaller.

%e For n = 12, gcd(12, binomial(12, 6)) = gcd(12, 924) = 12, so a(12) = 1.

%p swing := n -> n!/iquo(n,2)!^2: seq(n/igcd(n,swing(n)),n=1..92); # _Peter Luschny_, May 16 2013

%t Flatten[Table[{1, n/GCD[n, Binomial[n, n/2]]}, {n, 2, 100, 2}]] (* _Alonso del Arte_, May 17 2013 *)

%o (PARI) a(n) = n/gcd(n, binomial(n, n\2)); \\ _Michel Marcus_, Mar 22 2020

%Y Cf. A001405, A057815, A014973.

%K nonn

%O 1,4

%A _Labos Elemer_, Nov 13 2000

%E Offset changed to 1 by _Peter Luschny_, May 16 2013