Question on sum of digits of squares

Question

Let $D$ be the function define as $D(b,n)$ be the sum of the base-$b$ digits of $n$.

Example: $D(2,7)=3$ means $7=(111)_2\implies D(2,7)=1+1+1=3$

Define $S_m(a)=1^m+2^m+3^m+...+a^m$ where $a,m\in\mathbb{Z}_+$

Can it be shown that

(1)$$D(a,S_2(a))\le 2(a-1)?$$

(2) $$D(a,S_2(a))< a\iff a\equiv5\mod6?$$

Note: For $a,m>1$

● $a^m<S_m(a)<a^{m+1}$

● $1\le D(a,S_m(a))\le(a-1)(m+1)$

● $D(a,S_m(a))=1+D(a,S_m(a-1))$proof

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Edit

● $a\mid S_2(a)$ then $D(a+1,S_2(a+1))=a+1$

Proof:

let $b=a+1$.

Identically, we have $$ S_2(n) = \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ hence \begin{align*} &a{\,|\,}S_2(a)\\[4pt] \implies\;&a{\;|}\left( \frac{a(a+1)(2a+1)}{6} \right)\\[4pt] \implies\;&6{\;|}\left((a+1)(2a+1)\right)\\[4pt] \implies\;&6{\;|}\left(b(2b-1)\right)\\[4pt] \implies\;&6{\,|\,}b\;\;\text{or}\;\;\Bigl(2{\,|\,}b\;\;\text{and}\;\;3{\;|\,}(2b-1)\Bigr)\\[4pt] \end{align*} If $6{\,|\,}b$, then \begin{align*} S_2(b)&=\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 $$ D(b,S_2(b)) = \left({\small{\frac{b}{3}}}\right) + \left({\small{\frac{b}{2}}}\right) + \left({\small{\frac{b}{6}}}\right) = b $$ If $2{\,|\,}b\;\;$and$\;\;3{\;|\,}(2b-1)$, then $b\equiv 2\;(\text{mod}\;3)$, so \begin{align*} S_2(b)&=\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 $$ D(b,S_2(b)) = \left({\small{\frac{b+1}{3}}}\right) + \left({\small{\frac{b-2}{6}}}\right) + \left({\small{\frac{b}{6}}}\right) = b $$ Thus, for all cases, we have $D(b,S_2(b))=b$.

Answer 1

We can continue your casework on $ n \pmod{6}$.

Here's a sketch of it. Fill in the rest of the details yourself.

If $ n \equiv 0 \pmod{6}$, then $ \frac{ 2n^3 + 3n^2 + n } { 6} = \frac{2n}{6} \times n^2 + \frac{3n}{6} \times n + \frac{n}{6} \times 1 $,
so $D_2 = \frac{2n}{6} + \frac{3n}{6} + \frac{n}{6} = n $.
If $ n \equiv 1 \pmod{6}$, then $ \frac{ 2n^3 + 3n^2 + n }{6} = \frac{ 2n-2}{6} \times n^2 + \frac{ 5n - 5 } { 6} \times n + \frac{ n + 5 } { 6}\times 1 $,
so $D_2 = \frac{ 2n-2}{6} + \frac{ 5n-5}{6} + \frac{n+5}{6} = \frac{ 8n-2}{6}$.
If $ n\equiv 2 \pmod{6}$, then $ \frac{ 2n^3 + 3n^2 + n }{6} = \frac{2n - 4}{6} \times n^2 + \ldots$
so $D_2 = \ldots $
$\vdots$
If $ n \equiv 5 \pmod{6}$, then $ \frac{ 2n^3 + 3n^2 + n }{6} = \frac{2n+2}{6} \times n^2 + \frac{n+1}{6} \times n + 0 \times 1 $,
so $D_2 = \frac{2n+2}{6} + \frac{n+1}{6} = \frac{3n+3}{6} < n$.

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I have not done the rest as yet, but I believe it will all work out (assuming the statement is true). I'd be happy to review your algebra if you are unable to conclude that $D_2 \leq 2 (n-1)$ and $ D_2 < n \iff n\equiv 5 \pmod{6}$.