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Number of divisors k of n with gcd(k+1, n) = 1.
1

%I #12 Sep 07 2020 12:18:03

%S 1,1,2,2,2,1,2,3,3,2,2,3,2,2,3,4,2,2,2,3,4,2,2,4,3,2,4,4,2,3,2,5,3,2,

%T 4,5,2,2,4,5,2,1,2,4,4,2,2,6,3,3,3,4,2,3,4,6,4,2,2,5,2,2,6,6,4,3,2,4,

%U 3,3,2,7,2,2,5,4,4,2,2,7,5,2,2,5,4,2,3,6,2,5,3,4,4,2,3,7,2,3,5,5,2,3,2,6,6

%N Number of divisors k of n with gcd(k+1, n) = 1.

%e The positive divisors of 8 are 1, 2, 4 and 8. (2+1), (4+1) and (8+1) are relatively prime to 8, so a(8) = 3.

%t Table[Length[Select[Divisors[n],CoprimeQ[#+1,n]&]],{n,120}] (* _Harvey P. Dale_, Sep 07 2020 *)

%K nonn

%O 1,3

%A _Leroy Quet_, Aug 10 2000