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a(n) = 4^(2*n+1).
16

%I #51 Aug 26 2024 14:31:15

%S 4,64,1024,16384,262144,4194304,67108864,1073741824,17179869184,

%T 274877906944,4398046511104,70368744177664,1125899906842624,

%U 18014398509481984,288230376151711744,4611686018427387904,73786976294838206464,1180591620717411303424,18889465931478580854784

%N a(n) = 4^(2*n+1).

%C Also powers of 2 with singly even numbers (A016825) as exponents. - _Alonso del Arte_, Sep 03 2012

%C The partial sum of A000888(n) = Catalan(n)^2*(n + 1) resp. A267844(n) = Catalan(n)^2*(4n + 3) resp. A267987(n) = Catalan(n)^2*(4n + 4) divided by A013709(n) (this) a(n) = 2^(4n+2) absolutely converge to 1/Pi resp. 1 resp. 4/Pi. Thus this series is 1/Pi resp. 1 resp. 4/Pi. - _Ralf Steiner_, Jan 23 2016

%H Vincenzo Librandi, <a href="/A013709/b013709.txt">Table of n, a(n) for n = 0..200</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H <a href="/index/Di#divseq">Index to divisibility sequences</a>

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (16).

%F a(n) = 16*a(n-1), n > 0; a(0) = 4. G.f.: 4/(1 - 16*x). [_Philippe Deléham_, Nov 23 2008]

%F a(n) = 4^(2*n + 1) = 2^(4*n + 2). - _Alonso del Arte_, Sep 03 2012

%F a(n) = 4*A001025(n). - _Michel Marcus_, Jan 30 2016

%F From _Elmo R. Oliveira_, Aug 26 2024: (Start)

%F E.g.f.: 4*exp(16*x).

%F a(n) = A000302(A005408(n)). (End)

%p A013709:=n->4^(2*n+1): seq(A013709(n), n=0..20); # _Wesley Ivan Hurt_, Jan 30 2016

%t 2^(4 Range[0, 15] + 2) (* _Alonso del Arte_, Sep 03 2012 *)

%t NestList[16#&,4,20] (* _Harvey P. Dale_, Jun 03 2013 *)

%o (Magma) [4^(2*n+1): n in [0..20]]; // _Vincenzo Librandi_, May 26 2011

%o (PARI) a(n)=4<<(4*n) \\ _Charles R Greathouse IV_, Apr 07 2012

%Y Cf. A000888, A001025, A016825, A267844, A267987.

%Y Cf. A000302, A005408.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_