The sequence with k terms be (a1,a2),(a2,a3),...,(ak,ak+1) Where ai=2ai+1+1, using A on A relation a1=2a2+1,a1 will be odd. a2=2a3+1⇒a1=2(2a3+1)+1=4a3+3 a3=2a4+1a1=4(2a4+1)+3= ⋮ ak=2ak+1+1⇒a1=2k⋅ak+7 ak+1+(2k−1)∈A
a1+1−2k
2k
=ak+1⇒2krvert(a1+1), we need to find highest k.a1+1∈{2,...,101} k maximum when k=6, as at k=7,2k=128128∣ei∀ei∈A⇒a1=95 and k=6(95,47),(47,23),(23,11),(11,5),(5,2) will be sequence.
