Consider the sequence of numbers defined recursively by t1=1 and for n>1 by tn=1+t(n/2) when n is even and by tn=t(n−1)1 when n is odd. Given that tn=8719, the sum of the digits of n is<spanclass=′latex−bold′>(A)</span>15<spanclass=′latex−bold′>(B)</span>17<spanclass=′latex−bold′>(C)</span>19<spanclass=′latex−bold′>(D)</span>21<spanclass=′latex−bold′>(E)</span>23