%I #20 Aug 02 2024 17:00:01
%S 1,0,0,0,1,1,0,0,1,0,1,1,1,0,0,0,0,1,0,1,2,0,0,1,0,2,1,1,1,1,1,0,0,1,
%T 1,1,0,1,2,1,1,1,0,0,3,0,0,1,1,2,0,1,2,1,0,2,0,2,0,1,1,1,1,0,2,1,0,0,
%U 0,2,1,1,1,1,1,0,1,2,0,1,1,0,1,1,0,1,0,0,0,2,2,0,0,0,0,1,1,1,0,1,0,0,1,1,1
%N Number of values of s, 0 <= s <= n-1, such that 2^s mod n = s.
%H Michel Marcus, <a href="/A065293/b065293.txt">Table of n, a(n) for n = 1..1000</a>
%e For n=5 we have (2^0) mod 5 = 1, (2^1) mod 5 = 2, (2^2) mod 5 = 4, (2^3) mod 5 = 3, (2^4) mod 5 = 1. Only for s=3 does (2^s) mod 5=s, so a(5)=1
%t Table[Count[Range[0, n - 1], _?(Mod[2^#, n] == # &)], {n, 105}] (* _Michael De Vlieger_, Jun 19 2018 *)
%t Table[Count[Range[0,n-1],_?(PowerMod[2,#,n]==#&)],{n,110}] (* _Harvey P. Dale_, Aug 02 2024 *)
%o (PARI) a(n) = sum(s=0, n-1, Mod(2, n)^s == s); \\ _Michel Marcus_, Jun 19 2018
%Y Cf. A065294.
%K nonn
%O 1,21
%A Jonathan Ayres (jonathan.ayres(AT)btinternet.com), Oct 28 2001
%E a(1) corrected by _Michel Marcus_, Jun 20 2018