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%I #48 Sep 08 2022 08:44:38
%S 4,64,1024,16384,262144,4194304,67108864,1073741824,17179869184,
%T 274877906944,4398046511104,70368744177664,1125899906842624,
%U 18014398509481984,288230376151711744,4611686018427387904
%N a(n) = 4^(2n+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
%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.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_
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