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a(n) = ((3^p - 3)/p) mod p where p is n-th prime.
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%I #12 Nov 03 2024 19:26:15

%S 1,2,3,4,0,11,13,16,5,16,20,17,22,6,33,16,5,39,45,25,5,4,26,72,21,53,

%T 43,80,85,12,53,94,54,135,113,132,125,32,34,163,100,147,52,61,84,46,

%U 54,80,122,103,83,43,109,87,127,125,239,129,63,98,160,223,29,82,3,68,288,322

%N a(n) = ((3^p - 3)/p) mod p where p is n-th prime.

%C a(n) = 0 where n=5 (p=11) and n=78940 (p=1006003) see A014127.

%H Robert Israel, <a href="/A179078/b179078.txt">Table of n, a(n) for n = 1..10000</a>

%p f:= p -> (3&^p-3 mod p^2)/p:

%p seq(f(ithprime(i)),i=1..100); # _Robert Israel_, Nov 03 2024

%t aa = {}; Do[AppendTo[aa, Mod[(3^Prime[n] - 3)/Prime[n], Prime[n]]], {n, 1, 100}]; aa (* _Artur Jasinski_ *)

%Y Cf. A014127, A179077, A377669.

%K nonn

%O 1,2

%A _Artur Jasinski_, Jun 28 2010