Trying to prove a power sum divisibility conjecture For positive integers $n$ and $a$, define the integer power sum function
$$S_n(a) = 1^n+2^n+\dotsb+(a-1)^n+a^n\!.$$
I’m trying to prove

Conjecture: $(2a+3) \nmid S_{2n}(a)$

or, alternatively, find a counterexample.
Numeric calculations (a.k.a. brute-force searches) have turned up no counterexamples, and algebraic manipulation with “small” $n$ suggest that
$$(2a+3) \mid d_{2n}\bigl(2^{2n}S_{2n}(a)+1\bigr),$$
where $d_{2n}$ is the denominator of the power sum function (and thus related to Bernoulli numbers/polynomials); note that $d_{2n}$ is known to be squarefree.
By well-known [classical] results, we can deduce that $2a+3$ is squarefree, and that any prime $p$ dividing $2a+3$ must satisfy $(p-1) \mid 2n$. There are lots of recurrences out there which connect $S_{2n}(a)$ to $S_k(a)$ for various $k$ (especially $0 ≤ k < n$) — really, there is no shortage of literature on the integer power sum function. It seems like it should be easy to prove this conjecture… but after much work, I’m stuck.
Any suggestions — or a proof/disproof — would be appreciated.
 A: Not an answer, but maybe an explanation why counterexamples are sparse, if there are any.
\begin{align}
S_{2n}(a) &= 1^{2n}+2^{2n}+\dotsb+a^{2n} \\
&\equiv (2a+2)^{2n}+(2a+1)^{2n}+\cdots +(a+3)^{2n}\pmod{2a+3}.
\end{align}
Then:
$$S_{2n}(2a+2)\equiv 2S_{2n}(a)+(a+1)^{2n}+(a+2)^{2n}\pmod{2a+3}$$
Letting $m=2a+3,$ then $a+1=\frac{m-1}2,a+2=\frac{m+1}2,$ so you get:
$$2^{2n-1}S_{2n}(2a+2)\equiv 2^{2n}S_{2n}(a)+1\pmod m$$
(Is this true?) The closed formula of $S_{2n}(k)$ has $(k+1)$ in the numerator. When $k=2a+2,$ there will often be a prime divisor of $2a+3$ which is a divisor of $S_{2n}(2a+2).$ If they have a prime factor in common, then it isn’t possible for $2a+3\mid S_{2n}(a).$
For example, when $n=1,$ the only possible counter-example is $2a+3=3,$ because the denominator of the formula is $6.$
When $n=2,$ the only possible counterexamples are $2a+3\mid 15.$
In general, the finite set of counterexample possibilities are the $2a+3\mid (2n+1)!$

The only reason I think there might be a counterexample is that $a=0$ is always a counter-example, where $S_{2n}(0)=0.$ But the condition $a>0$ eliminates those.
A: Not an answer but some progress:
Consider @Thomas Andrews congruence
$$ 2^{2n-1}S_{2n}(2a+2) \equiv 2^{2n}S_{2n}(a)+1 \pmod {2a+3}$$
together with Carlitz-Von Staudt theorem  (L. Carlitz, The Staudt-Clausen theorem, Math. Mag. 34 (1960/1961), 131-146. Or see this paper,  for instance)
$$S_{2n}(2a+2)\equiv -\sum_{ p-1 \vert 2n\\\ p \vert 2a+3 }\frac{2a+3}{p }\pmod {2a+3}$$
where the sum is over those primes satisfying $p-1 \vert 2n$ and $p \vert 2a+3$ simultaneously.
This readily gives
$$\Big( 2^{2n}S_{2n}(a)+1\Big)\prod_{ p-1 \vert 2n\\\ p \vert 2a+3 }p \equiv 0 \pmod {2a+3}. $$
Now, if $2a+3$ divides $S_{2n}(a) $, we would have
$$\prod_{ p-1 \vert 2n\\\ p \vert 2a+3 }p \equiv 0 \pmod {2a+3} $$
which can only be true when $2a+3$ is squarefree and  $\underset{p\vert2a+3}{\text{lcm}}(p-1)$ divides $2n$.
But if these instances could be ruled out separately,  this would settle the conjecture.
Let  $q$ be any prime which divides the squarefree $2a+3$ and $n$ such that $\underset{q\vert2a+3}{\text{lcm}}(q-1)$  divides $2n$.  By Fermat little theorem, we have
$$ S_{2n}(a)\equiv \sum_{j=1}^a 1-\sum_{j=1}^a [q \vert j] \pmod q $$
$$ S_{2n}(a)\equiv a-  \lfloor \frac{a}{q}\rfloor \pmod q.$$
Let  $a = \lfloor \frac{a}{q}\rfloor\cdot q+r$ with $0\le r \le q-1$, then $2a+3 \equiv 2r+ 3 \bmod q$ and $ 3 \le 2r+3 \le 2q+1 $ and $q$ divide $2r+3$  then $2r +3 = q$. Then $2a+3 =\big(2 \lfloor \frac{a}{q}\rfloor+1\big)q$.
Then $a-\lfloor \frac{a}{q}\rfloor= \frac{2 q\lfloor \frac{a}{q}\rfloor+q-3}{2}-\lfloor \frac{a}{q}\rfloor= (q-1)\lfloor \frac{a}{q}\rfloor+\frac{q-3}{2} \equiv -\lfloor \frac{a}{q}\rfloor- \frac{3}{2} \bmod q $
Then a prime factor $q$ of squarefree $2a+3$ divides $S_{2n}(a) $ implies $q$ divides $2  \lfloor\frac{a}{q}\rfloor+3$.
Then if a square free $2a+3$ divides $S_{2n}(a)$,
it would also divide $\prod_{q\vert2a+3}\big(2  \lfloor\frac{a}{q}\rfloor+3\big)=\prod_{q\vert2a+3}\big(\frac{2a+3}{q}+2\big)$, since $2\lfloor\frac{a}{q}\rfloor= \frac{2a+3}{q}-1$.
This is the question (yet unanswered) raised here.
