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%I #26 Sep 21 2017 11:00:13
%S 0,1,3,1,15,1,63,1,255,1,1023,1,4095,1,16383,1,65535,1,262143,1,
%T 1048575,1,4194303,1,16777215,1,67108863,1,268435455,1,1073741823,1,
%U 4294967295,1,17179869183,1,68719476735,1,274877906943,1,1099511627775,1
%N Jacobsthal gap sequence.
%C Inverse binomial transform of A080925
%C From _Peter Bala_, Dec 26 2012: (Start)
%C Let F(x) = product {n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number F(1/4) = 0.79761 68651 30459 16010 ... = 1/(1 + 1/(3 + 1/(1 + 1/(15 + 1/(1 + 1/(63 + 1/(1 + 1/(255 + ...)))))))). See A111317. (End)
%C Also, the decimal representation of the diagonal from the corner to the origin of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 3", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - _Robert Price_, Apr 19 2017
%D S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.
%H Vincenzo Librandi, <a href="/A080924/b080924.txt">Table of n, a(n) for n = 0..300</a>
%H N. J. A. Sloane, <a href="http://arxiv.org/abs/1503.01168">On the Number of ON Cells in Cellular Automata</a>, arXiv:1503.01168 [math.CO], 2015
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ElementaryCellularAutomaton.html">Elementary Cellular Automaton</a>
%H S. Wolfram, <a href="http://wolframscience.com/">A New Kind of Science</a>
%H Wolfram Research, <a href="http://atlas.wolfram.com/">Wolfram Atlas of Simple Programs</a>
%H <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>
%H <a href="https://oeis.org/wiki/Index_to_2D_5-Neighbor_Cellular_Automata">Index to 2D 5-Neighbor Cellular Automata</a>
%H <a href="https://oeis.org/wiki/Index_to_Elementary_Cellular_Automata">Index to Elementary Cellular Automata</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-1,4,4).
%F a(2n) = 3*A001045(2n) = 3*A002450(n) = 4^n-1, a(2n+1)=1.
%F a(n) = (2^n-2*(-1)^n+(-2)^n)/2.
%F G.f.: x*(1+4*x)/((1+x)*(1+2*x)*(1-2*x)).
%F E.g.f.: (exp(2*x)-2*exp(-x)+exp(-2*x))/2.
%t CoefficientList[Series[x (1 + 4 x) / ((1 + x) (1 + 2 x) (1 - 2 x)), {x, 0, 50}], x] (* _Vincenzo Librandi_, Aug 05 2013 *)
%t LinearRecurrence[{-1, 4, 4}, {0, 1, 3}, 42] (* _Jean-François Alcover_, Sep 21 2017 *)
%Y Cf. A001045, A002450, A080926, A080927, A111317.
%K nonn,easy
%O 0,3
%A _Paul Barry_, Feb 26 2003