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a(n) = 2^n mod Fibonacci(n).
6

%I #22 Dec 07 2019 12:18:22

%S 0,0,0,1,2,0,11,4,2,34,1,64,37,173,438,394,118,1160,1663,1,6466,14508,

%T 20764,38368,18257,99928,64234,202972,15836,410224,184593,1520257,

%U 538006,2773540,5886173,9996832,5132559,9902536,21703576,33466456

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

%H Robert Israel, <a href="/A057862/b057862.txt">Table of n, a(n) for n = 1..4782</a>

%F a(n) = A000079(n) - A057861(n)*A000045(n)

%p seq(2 &^n mod combinat:-fibonacci(n), n=1..100); # _Robert Israel_, Jul 13 2018

%t Table[Mod[2^n,Fibonacci[n]],{n,1,80}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 03 2011*)

%t Table[PowerMod[2,n,Fibonacci[n]],{n,40}] (* _Harvey P. Dale_, Dec 30 2018 *)

%o (Sage) [power_mod(2,n,fibonacci(n))for n in range(1,41)] # _Zerinvary Lajos_, Nov 28 2009

%o (PARI) a(n)=2^n%fibonacci(n) \\ _Charles R Greathouse IV_, Jun 19 2017

%o (GAP) List([1..42],n->PowerMod(2,n,Fibonacci(n))); # _Muniru A Asiru_, Jul 13 2018

%Y Cf. A000045, A000079.

%K nonn

%O 1,5

%A _Henry Bottomley_, Sep 08 2000