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%I #7 Mar 30 2012 18:40:13
%S 1,2,7,207,700207,207000000700207,70020700000000000000207000000700207,
%T 2070000007002070000000000000000000000000000000000070020700000000000000207000000700207
%N a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the squares of the first n-1 terms of the sequence 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) = 14/2 = 7
%C ... 7^2 = 49 so a(4) = 1449/7 = 207
%C ... 207^2 = 42849 so a(5) = 144942849/207 = 700207.
%C For similarly defined sequences see A181754 through A181756 and A181865 through
%C A181870.
%F DEFINITION
%F a(1) = 1, a(2) = 2, and for n >= 3
%F (1)... a(n) = concatenate(a(1)^2,a(2)^2,...,a(n-1)^2)/a(n-1).
%F RECURRENCE RELATION
%F For n >= 2
%F (2)...a(n+2) = a(n+1) + 10^F(n,2)*a(n) = a(n+1) + 10^Pell(n)*a(n),
%F where F(n,2) is the Fibonacci polynomial F(n,x) evaluated at x = 2
%F and where Pell(n) = A000129(n).
%F RELATION WITH OTHER SEQUENCES
%F a(n) has A113225(n-2) digits.
%F a(n)^2 has Pell(n-1) digits.
%p #A181864
%p M:=8: a:=array(1..M):s:=array(1..M):
%p a[1]:=1:a[2]:=2:
%p s[1]:=convert(a[1]^2,string):
%p s[2]:=cat(s[1],convert(a[2]^2,string)):
%p for n from 3 to M do
%p a[n] := parse(s[n-1])/a[n-1];
%p s[n]:= cat(s[n-1],convert(a[n]^2,string));
%p end do:
%p seq(a[n],n = 1..M);
%Y A000129, A113225, A181754, A181755, A181756, A181865, A181866,A181867, A181868, A181869, A181870
%K nonn,easy
%O 1,2
%A _Peter Bala_, Nov 28 2010
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