2, 7, 15, 33,
3, 8, 16, 34, (add 1) ---- tn+1
6, 16, 32, 68, (double) ---- 2(tn+1)
6+1=7, 16-1=15, 32+1=33, (alternately add or subtract 1) ---- 2(tn+1)+(-1)n
So the next two terms in the sequence are 67 and 137.
Rule is: tn+1=2(tn+1)+(-1)n where t0=2; t1=2(2+1)+1=7; t2=2(7+1)-1=15; t3=2(15+1)+1=33; t4=2(33+1)-1=67; t5=2(67+1)+1=137.