cp's OEIS Frontend

A181869 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the squares of the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).

%I A181869 #4 Mar 30 2012 18:40:13
%S A181869 2,1,14,1401,140100014,140100014000000001401,
%T A181869 14010001400000000140100000000000000000000140100014
%N A181869 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the squares of the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).
%C A181869 The calculations for the first few values of the sequence are
%C A181869 ... 2^2 = 4 so a(3) = 14/1 = 14
%C A181869 ... 14^2 = 196 so a(4) = 19614/14 = 1401
%C A181869 ... 1401^2 = 1962801 so a(5) = 196280119614/1401 = 140100014
%C A181869 For similarly defined sequences see A181754 through A181756 and A181864 through A181870.
%F A181869 DEFINITION
%F A181869 a(1) = 2, a(2) = 1, and for n >= 3
%F A181869 (1)... a(n) = concatenate (a(n-1)^2,a(n-2)^2,...,a(1)^2)/a(n-1).
%F A181869 RECURRENCE RELATION
%F A181869 For n >= 2,
%F A181869 (2)... a(n+2) = 10^F(n,2)*a(n+1) + a(n) = 10^Pell(n)*a(n+1) + a(n),
%F A181869 where F(n,2) is the n-th Fibonacci polynomial F(n,x) evaluated at
%F A181869 x = 2, and Pell(n) = A000129(n).
%F A181869 a(n) has A024537(n-2) digits.
%p A181869 #A181869
%p A181869 M:=7:
%p A181869 a:=array(1..M):s:=array(1..M):
%p A181869 a[1]:=2:a[2]:=1:
%p A181869 s[1]:=convert(a[1]^2,string):
%p A181869 s[2]:=cat(convert(a[2]^2,string),s[1]):
%p A181869 for n from 3 to M do
%p A181869 a[n] := parse(s[n-1])/a[n-1];
%p A181869 s[n]:= cat(convert(a[n]^2,string),s[n-1]);
%p A181869 end do:
%p A181869 seq(a[n],n = 1..M);
%Y A181869 A000129, A024537, A181754, A181755, A181756, A181864, A181865, A181866, A181867, A181868, A181870
%K A181869 nonn,easy,base
%O A181869 1,1
%A A181869 _Peter Bala_, Nov 29 2010