Resisting Zero

Rank every number by its resistance, with the integer 1 being the sole number in row 1, integer 2 being the sole number in row 2, integers 3 and 4 being in row 3 and so on. Place row 1 at the bottom and work up.

Let $a(n)$ be the number of odd integers in row $n$ .

Let $b(n)$ be the number of even integers in row $n$ .

Let $c(n)$ be the total number of integers in row $n$ .

$c(n)=a(n)+b(n)$

Notice that an odd integer $p$ puts 1 integer (even) into the row above which is $2p$ .

An even integer $q$ puts two integers into the row above (one odd, one even) - $q+1$ and $2q$

Thus $a(n+1)=b(n)$

and

$b(n+1)=a(n)+b(n)$

$a(n+2)=b(n+1)$

and

$b(n+2)=a(n+1)+b(n+1)=a(n+1)+a(n)+b(n)$

$a(n+2)+b(n+2)=a(n+1)+b(n+1)+a(n)+b(n)$

$c(n+2)=c(n+1)+c(n)$

Stop right there! $c(n)$ , the number of integers in a row, is the Fibonacci sequence. 1,1,2,3,5,8,13

$c(7)=\boxed{13}$ .

If $R(n)$ is the resistance of $n$ then it is clear that $R(1) = 1$ and that $R(2n) \; = \; R(n) + 1 \hspace{2cm} R(2n+1) = R(n) + 2$ for all $n \in \mathbb{N}$ . This means that $R(n) \; = \; D(n) + C(n) - 1$ where $D(n)$ is the number of digits, and $C(n)$ the number of $1$ s, in the binary expansion of $n$ . To find numbers with resistance $7$ , we want numbers with $D(n) + C(n) = 8$ , and we have the following possibilities (we must have $D(n) \ge C(n) \ge 1$ ): $\begin{array}{|c|c|c|} \hline D(n) & C(n) & \mathrm{Count} \\ \hline 4 & 4 & 1 \\ 5 & 3 & \binom{4}{2} = 6 \\ 6 & 2 & \binom{5}{1} = 5 \\ 7 & 1 & \binom{6}{0} = 1 \\ \hline \end{array}$ (note that the first digit of a binary number must be $1$ ), making the total count $\boxed{13}$ .