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A238549 a(n) is one fourth of the total number of free ends of 4 line segments expansion at n iterations (see Comments lines for definition).

%I A238549 #60 Jul 04 2023 11:36:20
%S A238549 1,2,6,8,16,20,36,44,76,92,156,188,316,380,636,764,1276,1532,2556,
%T A238549 3068,5116,6140,10236,12284,20476,24572,40956,49148,81916,98300,
%U A238549 163836,196604,327676,393212,655356,786428,1310716,1572860,2621436,3145724,5242876,6291452,10485756
%N A238549 a(n) is one fourth of the total number of free ends of 4 line segments expansion at n iterations (see Comments lines for definition).
%C A238549 The initial pattern consists of 4 straight line segments which are the radii of a square. The next generations are scaled down by a factor of 1/sqrt(2) and rotated by an angle of Pi/4. Their free ends are the ends of elements that do not contact or cross the other ones. Overlaps among different generations are prohibited. See illustration in the links.
%C A238549 We take the official definition to be that provided by the PARI program. From this the assertions in the Formula section follow (they were formerly stated as conjectures). - _N. J. A. Sloane_, Feb 24 2019
%C A238549 From _Georg Fischer_, Feb 20 2019: (Start)
%C A238549 The following pattern can be seen for a(n) in base 2:
%C A238549    n         a(n)
%C A238549   ==  ==================
%C A238549    1    1 =          1_2
%C A238549    2    2 =         10_2
%C A238549    3    6 =        110_2
%C A238549    4    8 =       1000_2
%C A238549    5   16 =      10000_2
%C A238549    6   20 =      10100_2
%C A238549    7   36 =     100100_2
%C A238549    8   44 =     101100_2
%C A238549    9   76 =    1001100_2
%C A238549   10   92 =    1011100_2
%C A238549   11  156 =   10011100_2
%C A238549   12  188 =   10111100_2
%C A238549   13  316 =  100111100_2
%C A238549   14  380 =  101111100_2
%C A238549   15  636 = 1001111100_2
%C A238549   16  764 = 1011111100_2
%C A238549 (End)
%H A238549 Kival Ngaokrajang, <a href="/A238549/a238549_3.pdf">Illustration of initial terms</a>
%H A238549 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1, 2, -2).
%F A238549 a(n) =  1 + Sum_{i=1..n-1} A143095(i).
%F A238549 G.f.: x*(2*x^2+x+1) / ((x-1)*(2*x^2-1)). - _Colin Barker_, May 02 2015
%F A238549 From _Georg Fischer_, Feb 20 2019: (Start)
%F A238549 With p = floor((n + 2) / 2) for n >= 4: if n even then a(n) = 2^p + 4 * (2^(p - 4) - 1); if n odd then a(n) = 2^p + 4 * (2^(p - 3) - 1).
%F A238549 a(n) = a(n - 1) + 2 * a(n - 2) - 2 * a(n - 3).
%F A238549 (End)
%e A238549 The first numbers of free ends (4*a(n)) are 4, 8, 24, 32, 64, 80, 144, 176, 304, 368, 624, ...
%o A238549 (PARI) {print1(1,", "); for (n=1,100,s=1; for (i=0,n-1,s=s+(5-3*(-1)^i)*2^(1/4*(2*i-1+(-1)^i))/2); print1(s,", "))}
%o A238549 (Sage)
%o A238549 def a():
%o A238549     s, n = 2, 1
%o A238549     yield 1
%o A238549     while True:
%o A238549         yield s
%o A238549         s += (5-3*(-1)^n)*2^((2*n-1+(-1)^n)//4)//2
%o A238549         n += 1
%o A238549 A238549 = a(); [next(A238549) for _ in range(43)] # _Peter Luschny_, Feb 24 2019
%Y A238549 Cf. A143095, A256641.
%K A238549 nonn
%O A238549 1,2
%A A238549 _Kival Ngaokrajang_, May 01 2015