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Number of solutions to m^(1 + 2^v(n-1)) == -m (mod n), where v(n) = A007814(n) is the 2-adic valuation of n, and 0 <= m < n.
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%I #9 Feb 11 2023 08:09:31

%S 2,1,2,1,4,1,2,1,4,1,4,1,4,3,2,1,4,1,4,1,4,1,4,1,4,1,4,1,8,1,2,1,4,3,

%T 4,1,4,3,4,1,8,1,4,1,4,1,4,1,4,3,4,1,4,3,4,1,4,1,8,1,4,1,2,1,8,1,4,1,

%U 8,1,4,1,4,3,4,1,8,1,4,1,4,1,8,5,4,3,4,1,8,3,4,1,4,3,4,1,4,1,4,1,8,1,4,1,4,1,4,1,8,3,4,1,8,3,4,1,4,3,8,1

%N Number of solutions to m^(1 + 2^v(n-1)) == -m (mod n), where v(n) = A007814(n) is the 2-adic valuation of n, and 0 <= m < n.

%H Antti Karttunen, <a href="/A360112/b360112.txt">Table of n, a(n) for n = 2..16385</a>

%o (PARI) A360112(n) = { my(f=factor(n), x = 1+(2^valuation(n-1,2))); sum(m=0,n-1,!((m + m^x)%n)); };

%Y Cf. A007814, A345330 (composite numbers k, for which a(k) = 1), A345331 (odd numbers k, for which a(k) > 1), A360113, A360114 (positions of 1's).

%K nonn

%O 2,1

%A _Antti Karttunen_, Feb 10 2023