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).

1, 2, 6, 8, 16, 20, 36, 44, 76, 92, 156, 188, 316, 380, 636, 764, 1276, 1532, 2556, 3068, 5116, 6140, 10236, 12284, 20476, 24572, 40956, 49148, 81916, 98300, 163836, 196604, 327676, 393212, 655356, 786428, 1310716, 1572860, 2621436, 3145724, 5242876, 6291452, 10485756

Offset: 1

Author

Kival Ngaokrajang, May 01 2015

Keywords

• nonn

Comments

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.

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

From Georg Fischer, Feb 20 2019: (Start)

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

n a(n)

== ==================

1 1 = 1_2

2 2 = 10_2

3 6 = 110_2

4 8 = 1000_2

5 16 = 10000_2

6 20 = 10100_2

7 36 = 100100_2

8 44 = 101100_2

9 76 = 1001100_2

10 92 = 1011100_2

11 156 = 10011100_2

12 188 = 10111100_2

13 316 = 100111100_2

14 380 = 101111100_2

15 636 = 1001111100_2

16 764 = 1011111100_2

(End)

Examples

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

Links

• Kival Ngaokrajang, Illustration of initial terms
• Index entries for linear recurrences with constant coefficients, signature (1, 2, -2).

Crossrefs

Cf. A143095, A256641.

Programs

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,", "))}

Sage

def a():
    s, n = 2, 1
    yield 1
    while True:
        yield s
        s += (5-3*(-1)^n)*2^((2*n-1+(-1)^n)//4)//2
        n += 1
A238549 = a(); [next(A238549) for  in range(43)] # _Peter Luschny, Feb 24 2019

Formula

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

G.f.: x*(2*x^2+x+1) / ((x-1)*(2*x^2-1)). - Colin Barker, May 02 2015

From Georg Fischer, Feb 20 2019: (Start)

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).

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

(End)