OFFSET
1,2
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)
LINKS
Kival Ngaokrajang, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (1, 2, -2).
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)
EXAMPLE
The first numbers of free ends (4*a(n)) are 4, 8, 24, 32, 64, 80, 144, 176, 304, 368, 624, ...
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Kival Ngaokrajang, May 01 2015
STATUS
approved