%I #6 Mar 30 2012 18:40:13

%S 10,1,110,1001,1000110,10000001001,100000000001000110,

%T 10000000000000000010000001001,

%U 10000000000000000000000000000100000000001000110,1000000000000000000000000000000000000000000000010000000000000000010000001001

%N a(1) = 10, a(2) = 1. For n >= 3, a(n) = concatenate a(n-1), a(n-2), ..., a(1) and then divide the resulting number by a(n-1).

%C Compare with A181756.

%C The calculations for the first few values of the sequence are

%C ... a(3) = 110/1 = 110

%C ... a(4) = 110110/110 = 1001

%C ... a(5) = 1001110110 /1001 = 1000110

%C ... a(6) = 10001101001110110/1000110 = 10000001001.

%C The above calculations are in base 10, but we get exactly the same results when working in an arbitrary base b with initial values a(1) = 10 (= b in base b), a(2) = 1.

%C For similarly defined sequences see A181754 through A181756 and A181864 through A181870.

%F a(n+2) = a(n) + 10^(A000032(n+1)-1), n>=1.

%F a(n) has A000032(n-1) digits.

%p #A181868

%p M:=10:

%p a:=array(1..M):s:=array(1..M):

%p a[1]:=10:a[2]:=1:

%p s[1]:=convert(a[1],string):

%p s[2]:=cat(convert(a[2],string),s[1]):

%p for n from 3 to M do

%p a[n] := parse(s[n-1])/a[n-1];

%p s[n]:= cat(convert(a[n],string),s[n-1]);

%p end do:

%p seq(a[n],n = 1..M);

%Y A000032, A181754 - A181756, A181864 - A181870

%K nonn,easy

%O 1,1

%A _Peter Bala_, Nov 28 2010