
COMMENTS

It is easier to explain the rule of recurrence when the numbers are written as follows:
1,
2, 1,
5, 2, 2, 1,
20, 2, 5, 2, 5, 2, 2, 1,
95, 2, 5, 2, 20, 2, 5, 2, 20, 2, 5, 2, 5, 2, 2, 1,
470, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 20, 2, 5, 2, 5, 2, 2, 1,
2345, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 470, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 470,
2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 20, 2, 5, 2, 5, 2, 2, 1.
At first a(2^(n+1)1) = (3*5^n+5)/4 (n>=0). Let A be the sequence defined as follows:
A(0)=2; W(A(0))=5; A(1)=A(0),W(A(0))=2, 5; W(A(1))=2, 20.
More generally with A(n)=B(n), {3*5^n+5)/4; we define W(A(n))=B(n), (3*5^(n+1)+5)/4 and A(n+1)=A(n), W(A(n)).
Here we obtain A(1)=2, 5; W(A(1))=2, 20; A(2)=2, 5, 2, 20; W(A(2))=2, 5, 2, 95; A(3)=2, 5, 2, 20, 2, 5, 2, 95;
W(A(3))=2, 5, 2, 20, 2, 5, 2, 470; A(4)=2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, 470, etc.
In fact: B(1)=2; B(2)=2, 5, 2; B(3)=2, 5, 2, 20, 2, 5, 2; B(4)=2, 5, 2, 20, 2, 5, 2, 95, 2, 5, 2, 20, 2, 5, 2, etc.
If we denote by <<AUA>> the subsequence of a between a(2^(n+1)1) and a(2^(n+2)1), the subsequence of a between a(2^(n+2)1) and a(2^(n+3)1) is given by <<AA(n+1), A(n+1), UA>>.
It seems that this sequence gives the numbers of 1 in the successive sets of 1 in the sequence A174835.
