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a(n) = 2*A131577(n).
10

%I #27 Sep 05 2024 16:38:57

%S 0,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,

%T 131072,262144,524288,1048576,2097152,4194304,8388608,16777216,

%U 33554432,67108864,134217728,268435456,536870912,1073741824,2147483648,4294967296,8589934592

%N a(n) = 2*A131577(n).

%C Essentially the same as A131577, A046055, A011782, A000079 and A034008.

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (2).

%F a(n) = A000079(n), n>0.

%F a(n) = (-1)^(n+1)*A084633(n+1).

%F a(n) + A155543(n) = 2^n+4^n = A063376(n) = 2*A007582(n) =2*A137173(2n+1).

%F Conjecture: a(n) = A090129(n+3)-A090129(n+2).

%F G.f.: 2*x/(1-2*x). - _R. J. Mathar_, Jul 23 2009

%t CoefficientList[ Series[ 2x/(1 - 2x), {x, 0, 32}], x] (* _Robert G. Wilson v_, Aug 08 2018 *)

%o (PARI) a(n)=if(n,2^n,0) \\ _Charles R Greathouse IV_, Aug 01 2016

%o (Python)

%o def A155559(n): return 1<<n if n else 0 # _Chai Wah Wu_, Sep 05 2024

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Jan 24 2009

%E Edited by _R. J. Mathar_, Jul 23 2009

%E Extended by _Omar E. Pol_, Nov 19 2012