%I #18 Apr 26 2021 13:54:49

%S 1,10,110,10110,11010110,1011011010110,110101101011011010110,

%T 1011011010110110101101011011010110,

%U 1101011010110110101101011011010110110101101011011010110,10110110101101101011010110110101101101011010110110101101011011010110110101101011011010110

%N a(n) = (a(n-2) concatenate a(n-1)) for n > 2, with a(1)=1, a(2)=10.

%C Number of digits in a(n) = A000045(n+1). - _Michael S. Branicky_, Apr 24 2021

%C a(n) and a(n+1) contain Fibonacci(n) 1's and Fibonacci(n) 0's respectively.

%o (Python)

%o def aupton(terms):

%o alst = [1, 10]

%o for n in range(3, terms+1): alst.append(int(str(alst[-2])+str(alst[-1])))

%o return alst[:terms]

%o print(aupton(10)) # _Michael S. Branicky_, Apr 24 2021

%Y Cf. A000045, A111061 (in decimal), A061107, A131293.

%K nonn,base

%O 1,2

%A _Wesley Ivan Hurt_, Apr 24 2021