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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).
9
2, 1, 14, 1401, 140100014, 140100014000000001401, 14010001400000000140100000000000000000000140100014
OFFSET
1,1
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.
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.
MAPLE
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);
KEYWORD
nonn,easy,base
AUTHOR
Peter Bala, Nov 29 2010
STATUS
approved