cp's OEIS Frontend

A181867 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).

%I A181867 #4 Mar 30 2012 18:40:13
%S A181867 2,1,12,101,10012,10000101,1000000010012,100000000000010000101,
%T A181867 1000000000000000000001000000010012,
%U A181867 1000000000000000000000000000000000100000000000010000101
%N A181867 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).
%C A181867 Compare with A181754. Here we concatenate the terms of the sequence in reverse order before dividing by a(n-1).
%C A181867 The calculations for the first few values of the sequence are
%C A181867 ... a(3) = 12/1 = 12
%C A181867 ... a(4) = 1212/12 = 101
%C A181867 ... a(5) = 1011212/101 = 10012
%C A181867 ... a(6) = 100121011212/10012 = 10000101.
%C A181867 For similarly defined sequences see A181864 through A181870.
%F A181867 DEFINITION
%F A181867 a(1) = 2, a(2) = 1, and for n >= 3
%F A181867 (1)... a(n) = concatenate(a(n-1),a(n-2),...,a(1))/a(n-1).
%F A181867 RECURRENCE RELATION
%F A181867 For n >= 2
%F A181867 (2)... a(n+2) = a(n) + 10^(F(n)-1),
%F A181867 where F(n) = A000045(n) are the Fibonacci numbers.
%F A181867 a(n) has F(n) digits.
%p A181867 #A181867
%p A181867 M:=10:
%p A181867 a:=array(1..M):s:=array(1..M):
%p A181867 a[1]:=2:a[2]:=1:
%p A181867 s[1]:=convert(a[1],string):
%p A181867 s[2]:=cat(convert(a[2],string),s[1]):
%p A181867 for n from 3 to M do
%p A181867 a[n] := parse(s[n-1])/a[n-1];
%p A181867 s[n]:= cat(convert(a[n],string),s[n-1]);
%p A181867 end do:
%p A181867 seq(a[n],n = 1..M);
%Y A181867 A000045, A181754, A181755, A181756, A181864, A181865, A181866, A181868, A181869, A181870
%K A181867 nonn,easy,base
%O A181867 1,1
%A A181867 _Peter Bala_, Nov 28 2010