Prove that $W(n,\sum_{k=1}^{n}k^{2})=2(n-1)$

Let \begin{align} &a=n& &b=\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6 \end{align} We have $$\frac{a^3}3\leq b\leq\frac{a^3}3+a^2$$ from which $$2<\frac{\log(b)}{\log(a)}<3$$ hence $\lceil\log(b)/\log(a)\rceil=3$ and $W(a,b)=F_a(a^3-b)$. Moreover, if $b=d_0+d_1a+d_2a^2$ with $0\leq d_i\leq a-1$ then $$a^3-b=(a-d_0)+(a-d_1-1)a+(a-d_2-1)a^2$$ Since $a\nmid b$, we have $d_0\neq 0$, hence \begin{align} W(a,b) &=(a-d_0)+(a-d_1-1)+(a-d_2-1)\\ &=3a-(d_0+d_1+d_2)-2\\ &=3n-2-F_n(b) \end{align} and it remains to show $F_n(b)=n$.

If $n=6q$, then $b=q+(3q)n+(2q)n^2$, hence $F_n(b)=q+3q+2q=6q=n$.

If $n=6q+2$, then $b=(1+3q)+qn+(1+2q)n^2$, hence $F_n(b)=(1+3q)+q+(1+2q)=2+6q=n$.

For positive integers $n,k$, let $$S(n,k)=\sum_{i=1}^{n}i^k$$ and for positive integers $m,b$, with $b>1$, let $D(b,m)$ be the sum of the base-$b$ digits of $m$.

Let $k=2$.

Thus, suppose $a$ is a positive integer such that $a \mid S(a,2)$, and let $b=a+1$.

Identically, we have $$ S(n,2) = \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ hence \begin{align*} &a \mid S(a,2)\\[4pt] \implies\;&a{\;|}\left( \frac{a(a+1)(2a+1)}{6} \right)\\[4pt] \implies\;&6 \mid \left((a+1)(2a+1)\right)\\[4pt] \implies\;&6 \mid \left(b(2b-1)\right)\\[4pt] \implies\;&6 \mid b\;\;\text{or}\;\;\Bigl(2 \mid b\;\;\text{and}\;\;3 \mid (2b-1)\Bigr)\\[4pt] \end{align*} If $6 \mid b$, then \begin{align*} S(b,2)&=\frac{b(b+1)(2b+1)}{6}\\[4pt] &=\frac{b^3}{3}+\frac{b^2}{2}+\frac{b}{6}\\[4pt] &= \left({\small{\frac{b}{3}}}\right)\!{\cdot}\,b^2 + \left({\small{\frac{b}{2}}}\right)\!{\cdot}\,b^1 + \left({\small{\frac{b}{6}}}\right)\!{\cdot}\,b^0 \end{align*} hence $$ D(b,S(b,2)) = \left({\small{\frac{b}{3}}}\right) + \left({\small{\frac{b}{2}}}\right) + \left({\small{\frac{b}{6}}}\right) = b $$ If $2 \mid b$ and $3 \mid (2b-1)$, then $b\equiv 2 \pmod3$, so \begin{align*} S(b,2)&=\frac{b(b+1)(2b+1)}{6}\\[4pt] &=\frac{b^3}{3}+\frac{b^2}{2}+\frac{b}{6}\\[4pt] &= \left({\small{\frac{b+1}{3}}}\right)\!{\cdot}\,b^2 + \left({\small{\frac{b-2}{6}}}\right)\!{\cdot}\,b^1 + \left({\small{\frac{b}{2}}}\right)\!{\cdot}\,b^0 \end{align*} hence $$ D(b,S(b,2)) = \left({\small{\frac{b+1}{3}}}\right) + \left({\small{\frac{b-2}{6}}}\right) + \left({\small{\frac{b}{6}}}\right) = b. $$ Thus, for all subcases, we have $D(b,S(b,2))=b$.

So we are considering the summation $\sum_{i=1}^a i^2$ in base $a$. A quick observation yields $$ a^{2+1} = \sum_{i=1}^a a^2 \geq \sum_{i=1}^a i^2 \geq a^2. $$

See $W(b,S(b,2))=D(b,b^3-S(b,2))$

$\implies W(b,S(b,2))=3a+1-D(b,S(b,2))=2a$

and also note $a\in \{6t\pm 1\}$ then $a|S(a,2)$