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).

2, 1, 14, 1401, 140100014, 140100014000000001401, 14010001400000000140100000000000000000000140100014

Offset: 1

Author

Peter Bala, Nov 29 2010

Keywords

• nonn
• easy
• base

Comments

The calculations for the first few values of the sequence are

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

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

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

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

Crossrefs

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

Programs

Maple

#A181869
M:=7:
a:=array(1..M):s:=array(1..M):
a[1]:=2:a[2]:=1:
s[1]:=convert(a[1]^2,string):
s[2]:=cat(convert(a[2]^2,string),s[1]):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(convert(a[n]^2,string),s[n-1]);
end do:
seq(a[n],n = 1..M);

Formula

DEFINITION

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

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

RECURRENCE RELATION

For n >= 2,

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

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

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

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