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Jun 12 at 7:01 comment added noodle person Nice, that makes sense, thank you for explaining.
Jun 12 at 0:06 comment added Ka Wa Yip @noodleperson \$T_k\$ always goes up as \$k\$ increases (since one just keeping adding \$1/2\$ to it), while \$N\$ is given. One can choose a \$k\$ to satisfy \$T_k > N\$: for example, \$T_{2^{2N}} > N\$ . Let's say \$N = 3/2\$. \$2^{2(3/2)}=8\$. \$T_{8} = 1 + 1/2 +1/2+1/2 = 1+3/2 > 3/2\$. Since the input \$N\$ is an integer, a equivalent way to think about it is \$T_{2^{N}} > N/2\$. This is true because of the extra "1" at the beginning of \$T_k\$.
Jun 11 at 21:37 comment added noodle person Welcome to CGCC, nice first submission, and clever way of approaching this! I see \$ T_k > N \$ implies \$ H_k > N \$ but what if \$ H_k > N > T_k \$? Or does that case just never happen in this sequence?
S Jun 11 at 20:36 review First answers
Jun 12 at 10:57
S Jun 11 at 20:36 history answered Ka Wa Yip CC BY-SA 4.0