Sure. I did mention "x\\ne2^n" for positive innnteger "n". So, more extended:
"h(x)=\\begin{cases}\n \\lfloor\\log_2 x\\rfloor &\\text{if } x\\ne2^n \\\\\n (log_2 x)+1 &\\text{if } x=2^n\n\\end{cases}"for any arbitrary positive integer "n".
It can be seen inserting print(m) in the while loop.
So the answer is 12,
Comments
Leave a comment