For n > 4 it appears that a(n) = 4a(n1) + 4.
From Robert Israel, Nov 08 2016: (Start)
By induction, we have for k >= 0:
A075825((10*4^k7)/3) = 2^(k+1)
A075825((10*4^k4)/3) = 1
A075825((10*4^k1)/3) = 2^(k+1)+1
A075825((20*4^k8)/3) = 2^(k+1)1
A075825((20*4^k5)/3) = 2^(k+1)+1
A075825((20*4^k1)/3) = 2^(k+1)
In particular, this sequence contains b(k) = (10*4^k4)/3 which is the solution of b(k) = 4*b(k1)+4 with b(0) = 2.
The only terms <= 2*10^7 that are not of that form are 0 and 4. (End)
