%I #60 Jul 04 2023 11:36:20

%S 1,2,6,8,16,20,36,44,76,92,156,188,316,380,636,764,1276,1532,2556,

%T 3068,5116,6140,10236,12284,20476,24572,40956,49148,81916,98300,

%U 163836,196604,327676,393212,655356,786428,1310716,1572860,2621436,3145724,5242876,6291452,10485756

%N 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 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 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 From _Georg Fischer_, Feb 20 2019: (Start)

%C The following pattern can be seen for a(n) in base 2:

%C n a(n)

%C == ==================

%C 1 1 = 1_2

%C 2 2 = 10_2

%C 3 6 = 110_2

%C 4 8 = 1000_2

%C 5 16 = 10000_2

%C 6 20 = 10100_2

%C 7 36 = 100100_2

%C 8 44 = 101100_2

%C 9 76 = 1001100_2

%C 10 92 = 1011100_2

%C 11 156 = 10011100_2

%C 12 188 = 10111100_2

%C 13 316 = 100111100_2

%C 14 380 = 101111100_2

%C 15 636 = 1001111100_2

%C 16 764 = 1011111100_2

%C (End)

%H Kival Ngaokrajang, <a href="/A238549/a238549_3.pdf">Illustration of initial terms</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1, 2, -2).

%F a(n) = 1 + Sum_{i=1..n-1} A143095(i).

%F G.f.: x*(2*x^2+x+1) / ((x-1)*(2*x^2-1)). - _Colin Barker_, May 02 2015

%F From _Georg Fischer_, Feb 20 2019: (Start)

%F 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 a(n) = a(n - 1) + 2 * a(n - 2) - 2 * a(n - 3).

%F (End)

%e The first numbers of free ends (4*a(n)) are 4, 8, 24, 32, 64, 80, 144, 176, 304, 368, 624, ...

%o (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 (Sage)

%o def a():

%o s, n = 2, 1

%o yield 1

%o while True:

%o yield s

%o s += (5-3*(-1)^n)*2^((2*n-1+(-1)^n)//4)//2

%o n += 1

%o A238549 = a(); [next(A238549) for _ in range(43)] # _Peter Luschny_, Feb 24 2019

%Y Cf. A143095, A256641.

%K nonn

%O 1,2

%A _Kival Ngaokrajang_, May 01 2015