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a(n) = (-2)^n mod n.
3

%I #27 Mar 23 2020 06:22:24

%S 0,0,1,0,3,4,5,0,1,4,9,4,11,4,7,0,15,10,17,16,13,4,21,16,18,4,1,16,27,

%T 4,29,0,25,4,17,28,35,4,31,16,39,22,41,16,28,4,45,16,19,24,43,16,51,

%U 28,12,32,49,4,57,16,59,4,55,0,33,64,65,16,61,44,69,64,71,4,7

%N a(n) = (-2)^n mod n.

%C n^(n+2) mod (n+2) is essentially the same.

%C Indices of 0's: 2^k - 1, k>=0.

%C Indices of 1's: A006521 except the first term.

%C Indices of 3's: A015940.

%C Indices of 5's: 7, 133, 1517, 11761, ...

%C a(A000040(n)) = A000040(n)-2 = A040976(n).

%H Alois P. Heinz, <a href="/A215747/b215747.txt">Table of n, a(n) for n = 1..10000</a>

%e a(5) = (-2)^5 mod 5 = -32 mod 5 = 3.

%p a:= n-> (-2)&^n mod n:

%p seq(a(n), n=1..100); # _Alois P. Heinz_, Apr 08 2015

%o (Python)

%o for n in range(1, 333):

%o print((-2)**n % n, end=',')

%Y Cf. A015910, A082493, A082494, A110146, A213381.

%K nonn,look

%O 1,5

%A _Alex Ratushnyak_, Aug 23 2012