login
a(1) = 2; for n >= 2, n = Sum_{1<=k<n, gcd(k,n)=1} a(k).
1

%I #17 Jun 20 2024 08:36:07

%S 2,1,2,0,4,-2,0,2,6,-4,6,-4,-6,10,2,-2,12,-10,-2,8,2,-4,8,4,-16,8,10,

%T -8,10,-8,-8,14,14,-26,26,-14,-36,42,20,-22,68,-66,-60,14,-10,60,40,

%U -74,-38,-66,10,134,44,-98,-64,-54,22,156,20,-18,-34,-240,10,256,32,-18,-6,-144,-72,226,70,-68,-50,-184,58,236,82

%N a(1) = 2; for n >= 2, n = Sum_{1<=k<n, gcd(k,n)=1} a(k).

%H Sean A. Irvine, <a href="/A070963/b070963.txt">Table of n, a(n) for n = 1..10000</a>

%e 12 = a(1) + a(5) + a(7) + a(11) = 2 + 4 + 0 + 6 because 1, 5, 7 and 11 are the positive integers < 12 and relatively prime to 12.

%Y Cf. A045545.

%K sign

%O 1,1

%A _Leroy Quet_, May 16 2002