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

%I #15 Oct 30 2017 08:32:20

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

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

%U 3,4,1,8,1,2,4,4,3,4,1,7,4,2,1,6,3,2,3,6,1,4,3,4,2,2,3,8,1,3,5,6,1,3,1,6,4

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

%H Antti Karttunen, <a href="/A056692/b056692.txt">Table of n, a(n) for n = 1..16384</a>

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

%t Table[DivisorSum[n, 1 &, CoprimeQ[# - 1, n] &], {n, 105}] (* _Michael De Vlieger_, Oct 30 2017 *)

%o (PARI) A056692(n) = sumdiv(n,d,(1==gcd(d-1,n))); \\ _Antti Karttunen_, Oct 30 2017

%K nonn

%O 1,4

%A _Leroy Quet_, Aug 10 2000