Exponential Polynomial Summation!

We can derive properties of the polynomials $P_n$ by Lagrange interpolation. For any $0 \le x \le 2016$ , consider the degree $2016$ polynomial $\pi_x(X) \; = \; \prod_{{0 \le y \le 2016} \atop {y \neq x}}(X - x)$ noting that $\pi_x(y) = 0$ for integers $0 \le y \le 2016$ with $y \neq x$ . Then the desired polynomial $P_n(X)$ is $P_n(X) \; = \; \sum_{x=0}^{2016} \frac{n^x}{\pi_x(x)} \pi_x(X)$ and hence $P_n(2017) \; = \; \sum_{x=0}^{2016} \frac{n^x}{\pi_x(x)} \pi_x(2017)$ It is easy to calculate that $\pi_x(x) \; = \; (-1)^x \, x! \, (2016 - x)! \hspace{1cm} \pi_x(2017) \; = \; \frac{2017!}{2017 - x} \hspace{2cm} 0 \le x \le 2016$ and hence $\frac{\pi_x(2017)}{\pi_x(x)} \; =\; (-1)^x{2017 \choose x} \hspace{2cm} 0 \le x \le 2016$ so that $P_n(2017) \; = \; \sum_{x=0}^{2016} (-n)^x {2017 \choose x} \; = \; (1-n)^{2017} - (-n)^{2017} \; = \; n^{2017} - (n-1)^{2017}$ Thus we deduce that $\sum_{r=1}^{2017} P_r(2017) \; = \; 2017^{2017}$ making the answer $\lfloor 2017\log2017 \rfloor = \boxed{6665}$ .

Notice that

$\large \displaystyle (n+1) ^ x = \sum_{i=0}^{x} {x \choose i} n^i$ , where we define ${x \choose i}$ as a polynomial, where $\large { x \choose i} = \frac{ \displaystyle \prod_{j=0}^{i-1} (x-j)}{(i)!}$ and ${x \choose 0} = 1$ .

This inspires us to think of the polynomial $P_n$ in terms of combination polynomials, like the ones displayed above. Indeed,

$\large \displaystyle P_n(x) = \sum_{i=0}^{2016} (n-1)^i {x \choose i}$ is our needed $P_n$ (one can easily verify this by substituting values of $x$ ).

Also, we can see that

$\large P_n(2017) = \displaystyle \sum_{i=0}^{2016} (n-1)^i {2017 \choose i} = \sum_{i=0}^{2017} (n-1)^i {2017 \choose i} - (n-1)^{2017} = n^{2017} - (n-1)^{2017}$

(The second equality is from the fact that ${ 2017 \choose 2017} (n-1)^{2017} = (n-1)^{2017}$ ).

Thus, it follows that

$\large \displaystyle r = \sum_{i=1}^{2017} P_i(2017) = \sum_{i=1}^{2017} i^{2017} - (i-1)^{2017} = 2017^{2017}$

(Since the sum telescopes to $2017^{2017} - 0^{2017}$ ).

Our desired answer is then $\lfloor \log_{10} 2017^{2017} \rfloor = \lfloor 2017 \log _{10} 2017 \rfloor = \boxed{6665}$ .