A conjecture about the sum of Faulhaber polynomials mod 2.

Question

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 $f_2$ where $f_m(x)$ is apolynomial where $f_m(x)$ is odd for the first $m$ natural numbers and $f_m(x)$ is even for the second $m$ natural numbers and so on that made me do some experiments with $\sum\limits_{1\le k\le n} P_k(x)$.

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Let $P_n(x)$ be the $n-th$ Faulhaber polynomial let $Q_n(x)= \sum\limits_{1\le k\le n} P_k(x)$ 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):

$\begin{bmatrix} & \text{Even(E)} &\text{Odd (O)} \\S_{1} & 2& 2 \\ S_{2}& 3 &1 \\ S_{3}& 2&2 \\ S_{4}& 4&4 \\ S_{5}& 2&2 \\ S_{6}& 3&1 \\ S_{7}& 2&2 \\ S_{8}& 4&4 \\ S_{9}& 2&2 \\ S_{10}& 3&1 \\ S_{11}& 2&2 \\ S_{12}& 4&4 \end{bmatrix}$

It seems that $S_n$ will have a cycle $(2,2), \ (4,4), \ (2,2), \ (3,1)$ But I couldn't prove or disprove this, where $(E,O)$ determine how many numbers even numbers and odd numbers appear in each cycle respectfully.

It is easy to prove that for odd $n$, $(E,O)=(2,2)$ and $r(n)=0$.

Also $r_{2n}=1$ for all even numbers that I tried, for some reason $r_{2n}$ can't be greater than $1$.

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Here is the first twelve Faulhaber polynomials:

$$P_{1}(x)= \frac{x(x+1)}{2}$$ $$P_{2}(x)= \frac{x(2x+1)(x+1)}{6}$$ $$P_{3}(x)= \frac{x^2(x+1)^2}{4}$$ $$P_{4}(x)=\frac{x(2x+1)(x+1)}{6} \times \frac{3x^2+3x-1}{ 5}$$ $$P_{5}(x)= \frac{x^2(x+1)^2}{4} \times\frac{2x^2+2x-1}{ 3}$$ $$P_{6}(x)= \frac{x(2x+1)(x+1)}{6} \times \frac{3x^4+6x^3-3x+1}{ 7 }$$ $$P_{7}(x)= \frac{x^2(x+1)^2}{4 }\times\frac{3x^4+6x^3-x^2-4x+2}{ 6}$$ $$P_{8}(x)= \frac{x(2x+1)(x+1)}{6} \times \frac{5x^6+15x^5+5x^4-15x^3-x^2+9x-3}{ 15}$$ $$P_{9}(x)= \frac{x^2(x+1)^2}{4} \times\frac{(x^2+x-1)(2x^4 +4x^3-x^2-3x+3)}{ 5}$$ $$P_{10}(x)= \frac{x(2x+1)(x+1)}{6} \times \frac{ 3 x^8+ 12 x^7+ 8 x^6 - 18 x^5- 10 x^4+ 24 x^3 + 2 x^2 - 15 x +5}{ 11}$$ $$P_{11}(x)= \frac{x^2(x+1)^2}{4} \times\frac{2x^8 +8x^7+4x^6-16x^5-5x^4+26x^3-3x^2-20x+10}{ 6}$$ $$P_{12}(x)=\frac{x^{13}}{13}+\frac{x^{12}}{12}+x^{11}-\frac{11}{6}x^9+\frac{22x^7}{7}-\frac{33x^5}{10}+\frac{5x^3}{3}-\frac{961x}{2730}$$

Answer 1

Here, I assume $n$ is even (you have solved the for odd $n$). Also, since your claim assumes $x$ as a positive integer, we can simply use sums instead of Faulhaber's polynomials.

We need to show that $Q_n(x)\equiv 2 \pmod 4$ for some $x\in \mathbb{N}$ and $Q_n(x)\equiv 0 \pmod 2$ for all $x\in \mathbb{N}$ (so in your notation $r(n)=1$). This implies $S_n(x)=\frac{Q_n(x)}{2}$. As a result, we can then examine the sequence $Q_n(x) \bmod 4$ instead of $S_n(x) \bmod 2$.

First, we show $Q_n(x)\equiv 0 \pmod 2$. Since $x$ is a positive integer, we have $P_n(x)=\sum_{i=1}^{x}i^n$ and so

$$ Q_n(x)=\sum_{k=1}^{n} \sum_{i=1}^{x}i^k\equiv \sum_{k=1}^n \sum_{i=1}^xi=n\frac{x(x+1)}{2}\equiv 0\pmod 2. $$

Next, let's evaluate $Q_n(x)$ on first four values of $x$:

$$ Q_n(1)=n\\ Q_n(2)=\sum_{k=1}^{n}(1+2^k)=n+\sum_{k=1}^{n}2^k\equiv n+2 \pmod 4\\ Q_n(3)=\sum_{k=1}^{n}(1+2^k+3^k)\equiv n+2+\sum_{k=1}^{n}3^k\equiv n+2 \pmod 4\\ Q_n(4)=\sum_{k=1}^{n}(1+2^k+3^k+4^k)\equiv \sum_{k=1}^{n}(1+2^k+3^k)=Q_n(3)\equiv n+2 \pmod 4 $$ (we used that $\sum_{k=1}^{n}3^k$ is divisible by $4$ for even $n$, which is a simple exercise to prove).

For next values, the bases in the summands are just periodic modulo $4$, so

$$ Q_n(5)\equiv Q_n(4)+Q_n(1)\equiv 2n+2\equiv 2 \pmod 4\\ Q_n(6)\equiv Q_n(4)+Q_n(2)\equiv 2n+4\equiv 0 \pmod 4\\ Q_n(7)\equiv Q_n(4)+Q_n(3)\equiv 2n+4\equiv 0 \pmod 4\\ Q_n(8)\equiv Q_n(4)+Q_n(4)\equiv 2n+4\equiv 0 \pmod 4 $$

Notice that $Q_n(x)\equiv 2 \pmod 4$ for some values of $x$, finishing the proof that $S_n(x)=\frac{Q_n(x)}{2}$.

Now are are almost done. Notice that $Q_n(8)\equiv 0 \pmod 4$, and so the sequence $Q_n(x) \bmod 4$ is periodic with a length of $8$. Furthermore, the pattern is given by $$(n,n+2,n+2,n+2,2,0,0,0),$$ so it is either $(0,2,2,2,2,0,0,0)$ or $(2,0,0,0,2,0,0,0)$ for $n$ even, as required (former corresponds to four evens followed by four odds in $S_n(x)$, while latter corresponds to three evens followed by one odd).

Answer 2

Theorem 1 in a paper I wrote with Jonathan Sondow gives an elementary solution to the problem of determining the precise power of $2$ dividing any $S_n(a)$. I think you’ll find what you need using that.