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%I #21 Sep 08 2022 08:45:34
%S 4,13,49,193,769,3073,12289,49153,196609,786433,3145729,12582913,
%T 50331649,201326593,805306369,3221225473,12884901889,51539607553,
%U 206158430209,824633720833,3298534883329,13194139533313,52776558133249
%N a(n) = 3*4^n + 1.
%C An Engel expansion of 4/3 to the base 4 as defined in A181565, with the associated series expansion 4/3 = 4/4 + 4^2/(4*13) + 4^3/(4*13*49) + 4^4/(4*13*49*193) + .... Cf. A199115. - _Peter Bala_, Oct 29 2013
%H Vincenzo Librandi, <a href="/A140660/b140660.txt">Table of n, a(n) for n = 0..300</a>
%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-4).
%F a(n) = A002001(n+1) + 1.
%F a(n) = 4*a(n-1) - 3.
%F First differences: a(n+1) - a(n) = A002063(n).
%F a(n+k) - a(n) = 3*(4^k - 1)*A000302(n) = 9*A002450(k)*A000302(n).
%F a(n) = A140529(n) - A096045(n).
%F O.g.f.: (7*x - 4)/((1 - x)*(4*x - 1)). - _R. J. Mathar_, Jul 14 2008
%F From _G. C. Greubel_, Sep 15 2017: (Start)
%F E.g.f.: 3*exp(4*x) + exp(x).
%F a(n) = 5*a(n-1) - 4*a(n-2). (End)
%t LinearRecurrence[{5,-4}, {4,13}, 50] (* or *) CoefficientList[Series[ (7*x-4)/((1-x)*(4*x-1)), {x,0,50}], x] (* _G. C. Greubel_, Sep 15 2017 *)
%o (Magma) [3*4^n+1: n in [0..30] ]; // _Vincenzo Librandi_, May 23 2011
%o (PARI) x='x+O('x^50); Vec((7*x-4)/((1-x)*(4*x-1))) \\ _G. C. Greubel_, Sep 15 2017
%Y Cf. A181565, A199115.
%K nonn
%O 0,1
%A _Paul Curtz_, Jul 10 2008
%E Edited and extended _R. J. Mathar_, Jul 14 2008
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