Fibonacci LFSR, recovering previous bits
Yes. just run the lfsr backwards, so if you have $$s(t)=s(t-2)+s(t-7)$$ use $$s(t-7)=s(t-2)+s(t)$$ all mod 2 with $$(s(-6),\ldots,s(0))=(1,0,1,0,1,0,1)$$ your initial loading.