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A267796 a(n) = (n+1)*4^(2n+1).

%I A267796 #64 Apr 17 2022 08:21:30
%S A267796 4,128,3072,65536,1310720,25165824,469762048,8589934592,154618822656,
%T A267796 2748779069440,48378511622144,844424930131968,14636698788954112,
%U A267796 252201579132747776,4323455642275676160,73786976294838206464,1254378597012249509888,21250649172913403461632
%N A267796 a(n) = (n+1)*4^(2n+1).
%C A267796 The partial sums of A001246(n)/a(n) converge absolutely. This series is also the hypergeometric function 1/4 * 4F3(1/2,1/2,1,1;2,2,2;1). - _Ralf Steiner_, Feb 09 2016
%H A267796 Colin Barker, <a href="/A267796/b267796.txt">Table of n, a(n) for n = 0..800</a>
%H A267796 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (32,-256).
%F A267796 a(n) = A013709(n)*(n+1).
%F A267796 From _Colin Barker_, Mar 23 2017: (Start)
%F A267796 G.f.: 4 / (1 - 16*x)^2.
%F A267796 a(n) = 32*a(n-1) - 256*a(n-2) for n>1. (End)
%F A267796 From _Amiram Eldar_, Apr 17 2022: (Start)
%F A267796 a(n) = A193132(n+1)/3.
%F A267796 Sum_{n>=0} 1/a(n) = 4*log(16/15).
%F A267796 Sum_{n>=0} (-1)^n/a(n) = 4*log(17/16). (End)
%e A267796 For n=3, a(3) = (3+1)*4^(2*3+1) = 4*4^7 = 65536.
%t A267796 Table[(n + 1) 4^(2 n + 1), {n, 0, 20}] (* _Vincenzo Librandi_, Feb 10 2016 *)
%o A267796 (PARI) a(n) = (n+1)*4^(2*n+1); \\ _Michel Marcus_, Jan 28 2016
%o A267796 (PARI) Vec(4 / (1 - 16*x)^2 + O(x^30)) \\ _Colin Barker_, Mar 23 2017
%o A267796 (Magma) [(n+1)*4^(2*n+1): n in [0..45]]; // _Vincenzo Librandi_, Feb 10 2016
%Y A267796 Cf. A001246, A013709, A193132, A267982.
%K A267796 nonn,easy
%O A267796 0,1
%A A267796 _Ralf Steiner_, Jan 24 2016
%E A267796 More terms from _Michel Marcus_, Jan 28 2016