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A223925 a(2n+1) = 2*n-1; a(2n)= 4^n.

%I A223925 #37 Jul 30 2018 19:30:01
%S A223925 1,4,3,16,5,64,7,256,9,1024,11,4096,13,16384,15,65536,17,262144,19,
%T A223925 1048576,21,4194304,23,16777216,25,67108864,27,268435456,29,
%U A223925 1073741824,31
%N A223925 a(2n+1) = 2*n-1; a(2n)= 4^n.
%C A223925 If A132050(n) has offset 1 (proposed),
%C A223925 A132049(n)/A132050(n) = 2, 4, 3, 16/5, 25/8, 192/61,... leads to Pi (Euler, 1735)
%C A223925 A132049(n)/a(n) = (2/1=2, 4/4=1, 3/3=1, 16/16=1, 25/5=5, 192/64=3,... ). The second bisection  1, 1, 3, 17, 155, begins like A110501.
%C A223925 Conjecture: a(2n) is always a divisor of A132049(2n).
%H A223925 Vincenzo Librandi, <a href="/A223925/b223925.txt">Table of n, a(n) for n = 1..1000</a>
%H A223925 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,-9,0,4).
%F A223925 G.f.: x*(1+4*x-3*x^2-8*x^3-4*x^4+4*x^5)/((1-x)^2*(1+x)^2*(1-2x)*(1+2x)). - _Philippe Deléham_, Apr 01 2013
%F A223925 a(n) = 6*a(n-2) -9*a(n-4) + 4*a(n-6) with a(1) = 1, a(2) = 4, a(3) = 3, a(4) = 16, a(5) = 5, a(6) = 64. - _Philippe Deléham_, Apr 01 2013
%t A223925 Table[ If[ OddQ[n], n, 4^(n/2)], {n, 1, 31}] (* _Jean-François Alcover_, Apr 02 2013 *)
%t A223925 CoefficientList[Series[(1 + 4 x - 3 x^2 - 8 x^3 - 4 x^4 + 4 x^5) / ((1 - x)^2 (1 + x)^2 (1 - 2 x) (1 + 2 x)), {x, 0, 35}], x] (* _Vincenzo Librandi_, Jul 20 2013 *)
%t A223925 LinearRecurrence[{0,6,0,-9,0,4},{1,4,3,16,5,64},40] (* _Harvey P. Dale_, Jul 30 2018 *)
%Y A223925 Cf. A005408, A000302, A132050, A100102.
%K A223925 nonn,easy,less
%O A223925 1,2
%A A223925 _Paul Curtz_, Mar 29 2013
%E A223925 Conjecture about A132049(n)/a(n) modified by _Jean-François Alcover_, Apr 12 2013