Context: In this question I asked about "Finding a polynomial function for alternating $m$ numbers of odds and even numbers in sequences" I noticed that all Faulhaber polynomials are solutions for $f_2$ that made me do some experiments with $\sum\limits_{1\le k\le n} P_n(x)$
Let $P_n(x)$ be the $n-th$ Faulhaber polynomial let $Q_n(x)= \sum\limits_{1\le k\le n} P_n(x)$ let $r(n )\in \{0,1,2,3,\dots\}$ be the biggest number such that $\frac{Q_n(x)}{2^{r(n)}}$ is always integer when $x$ is a natural number Now define $S_n(x)= \frac{Q_n(x)}{2^{r(n)}}$. I noticed that $S_n(x)$ when $x\in \mathbb{N}$ always result on a block of $E$ even numbers followed by a block of $O$ odd numbers.for example $S_1(x)=\frac{x(x+1)}{2}$ when $x \in \mathbb{N}$ the sequence would be the following $1,3,6,10,15,21,28,36$ so $E=O=2$ (i.e 2 evens followed by 2 odds and the cycle continue)
The question is how to determine $E,O$ for any $S_n$?
Here is a table containing the first twelve of $S_n$ (This is not a matrix):
$\begin{bmatrix}& \text{Even(E)} &\text{Odd (O)}\\S_{1} & 2& 2\\ S_{2}& 2 &2\\ S_{3}& 2&2\\ S_{4}& 4&4\\ S_{5}& 2&2\\ S_{6}& 3&1\\ S_{7}& 2&2\\ S_{8}& 4&4\\ S_{9}& 2&2\\ S_{10}& 3&1\\ S_{11}& 2&2\\ S_{12}& 4&4\end{bmatrix}$
It seems that $S_n$ will have a cycle $(2,2), \ (4,4), \ (2,2), \ (3,1)$ But I couln't prove this where $(E,O)$ determine how many numbers even numbers and odd numbers appear in each cycle respectfully.
Here is the first twelve faulhaber polynomials
$$P_{1}(x)= \frac{x(x+1)}{2}$$$$P_{2}(x)= \frac{x(2x+1)(x+1)}{6}$$$$P_{3}(x)= \frac{x^2(x+1)^2}{4}$$$$P_{4}(x)=\frac{x(2x+1)(x+1)}{6} \times \frac{3x^2+3x-1}{ 5}$$$$P_{5}(x)= \frac{x^2(x+1)^2}{4} \times\frac{2x^2+2x-1}{ 3}$$$$P_{6}(x)= \frac{x(2x+1)(x+1)}{6} \times \frac{3x^4+6x^3-3x+1}{ 7 }$$$$P_{7}(x)= \frac{x^2(x+1)^2}{4 }\times\frac{3x^4+6x^3-x^2-4x+2}{ 6}$$$$P_{8}(x)= \frac{x(2x+1)(x+1)}{6} \times \frac{5x^6+15x^5+5x^4-15x^3-x^2+9x-3}{ 15}$$$$P_{9}(x)= \frac{x^2(x+1)^2}{4} \times\frac{(x^2+x-1)(2x^4 +4x^3-x^2-3x+3)}{ 5}$$$$P_{10}(x)= \frac{x(2x+1)(x+1)}{6} \times \frac{ 3 x^8+ 12 x^7+ 8 x^6 - 18 x^5- 10 x^4+ 24 x^3 + 2 x^2 - 15 x +5}{ 11}$$$$P_{11}(x)= \frac{x^2(x+1)^2}{4} \times\frac{2x^8 +8x^7+4x^6-16x^5-5x^4+26x^3-3x^2-20x+10}{ 6}$$$$P_{12}(x)=\frac{x^{13}}{13}+\frac{x^{12}}{12}+x^{11}-\frac{11}{6}x^9+\frac{22x^7}{7}-\frac{33x^5}{10}+\frac{5x^3}{3}-\frac{961x}{2730}$$