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a(n) = (2^n + 3^n) modulo n.
3

%I #11 Dec 26 2023 23:42:45

%S 0,1,2,1,0,1,5,1,8,3,5,1,5,13,5,1,5,1,5,17,14,13,5,1,0,13,26,13,5,13,

%T 5,1,2,13,30,1,5,13,35,17,5,37,5,9,35,13,5,1,12,23,35,45,5,1,0,41,35,

%U 13,5,37,5,13,35,1,15,1,5,29,35,13,5,1,5,13,50

%N a(n) = (2^n + 3^n) modulo n.

%C a(n) = (A015910(n) + A066601(n)) mod n.

%C a(n) = 0 at n = 1, 5, 25, 55, 125, 275, 605, 625, ... (A045576).

%H Zak Seidov, <a href="/A220235/b220235.txt">Table of n, a(n) for n = 1..1000</a>

%t Table[Mod[2^n + 3^n, n],{n,100}]

%Y Cf. A015910 (2^n mod n), A066601 (3^n mod n), A045576 (n|(2^n + 3^n)).

%K nonn

%O 1,3

%A _Zak Seidov_, Dec 08 2012