A181869 - OEIS

%I #4 Mar 30 2012 18:40:13

%S 2,1,14,1401,140100014,140100014000000001401,

%T 14010001400000000140100000000000000000000140100014

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

%C ... 2^2 = 4 so a(3) = 14/1 = 14

%C ... 14^2 = 196 so a(4) = 19614/14 = 1401

%C ... 1401^2 = 1962801 so a(5) = 196280119614/1401 = 140100014

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

%F DEFINITION

%F a(1) = 2, a(2) = 1, and for n >= 3

%F (1)... a(n) = concatenate (a(n-1)^2,a(n-2)^2,...,a(1)^2)/a(n-1).

%F RECURRENCE RELATION

%F For n >= 2,

%F (2)... a(n+2) = 10^F(n,2)*a(n+1) + a(n) = 10^Pell(n)*a(n+1) + a(n),

%F where F(n,2) is the n-th Fibonacci polynomial F(n,x) evaluated at

%F x = 2, and Pell(n) = A000129(n).

%F a(n) has A024537(n-2) digits.

%p #A181869

%p M:=7:

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

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

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

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

%p end do:

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

%Y A000129, A024537, A181754, A181755, A181756, A181864, A181865, A181866, A181867, A181868, A181870

%K nonn,easy,base

%O 1,1

%A _Peter Bala_, Nov 29 2010