How to prove this sum involving powers of cosec is an integer?

Question

It is claimed that the following function produces only integer values for all integer $m \geq 1$, $N \geq 2$.

$F(m,N)=\frac{N^m}{2^m}\displaystyle \sum_{j=1}^{N-1} \operatorname{cosec} ^{2m}\left(\frac{\pi j}{N}\right)$

How do I prove this?

(Source: https://twitter.com/SamuelGWalters/status/1513266704855363587 - and extensive testing numerically makes me reasonably confident it is true).

Edit to say that now I have found a solution, the following comments are not necessary to tackle the problem (but might provide further insight into F).

I found this paper: fq.math.ca/Papers1/44-3/quartgauthier03_2006.pdf. I used the method there to express $F$ solely in terms of rational numbers - in outline, this went as follows...

Writing $g_m(z) = \text{cosec}^{2m+2}(z)$ it can be shown that

$\frac{d^{2}}{dz^{2}}g_m(z) = (2m+2)(2m+3)g_{m+1}(z)-(2m+2)^2g_m(z)$

and so writing

$g_m(z)=\displaystyle\sum_{r=0}^m \phi_{r,m} \frac{d^{2r}}{dz^{2r}}[g_0(z)]$

we can establish that

$\phi_{r,m}=0$ for $r <0$ and $r>m$,

$\phi_{0,0}=1$,

$\phi_{r,m+1}=\large \frac{\phi_{r-1,m}+(2m+2)^2\phi_{r,m}}{(2m+2)(2m+3)}$

From the Mittag-Leffler theorem,

$g_0(z)=\displaystyle\sum_{n\in\mathbb{Z}}(z-n\pi)^{-2}$ so

$\frac{d^{2r}}{dz^{2r}}[g_0(z)]=(2r+1)!\displaystyle\sum_{n\in\mathbb{Z}}(z-n\pi)^{-2(r+1)}$

This enables us to write:

$F(m,N)=\displaystyle\sum_{r=1}^{m}(2r-1)!\phi_{r-1,m-1}T_{r,m}(N)$

where $T_{r,m}(N)=\displaystyle \frac{N^{m+2r}}{2^m\cdot\pi^{2r}}\displaystyle \sum_{j=1}^{N-1}\displaystyle\sum_{n\in\mathbb{Z}}( j-nN)^{-2r}$

Then using properties of the Riemann-zeta function, including for integer $r$ that $\zeta(2r)=2^{2r-1}|B_{2r}|\pi^{2r}/(2r)!$, where $(B_{2r})$ are Bernouilli numbers it can be shown that

$T_{r,m}(N) = 2^{2r-m}(N^{2r+m}-N^m)|B_{2r}|/(2r)!$

In summary,

$F(m,N)=\displaystyle\sum_{r=1}^{m}2^{2r-m-1} r^{-1} \phi_{r-1,m-1} |B_{2r}|N^m(N^r-1)(N^r+1)$

$\phi_{r,m}$ as defined above.

For reference, $B_0 = 1$, $B_1 = \frac12$, $B_r = 0$ for odd $r > 1$,

$\displaystyle \sum_{k=0}^m {m+1 \choose k}B_k=m+1$

How can I show $F(m,N)$ is always an integer for $m \geq 1, N \geq 2$?

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ADDITIONAL NOTE

For reference, by numerical grind, I have found these formulae (and confirmed they give integer values):

$F(1,N) = N(N-1)(N+1)/(2 \cdot 3)$

$F(2,N) = N^2 (N-1)(N+1)(N^2+11)/(2^2 \cdot 3^2 \cdot 5)$

$F(3,N) = N^3(N-1)(N+1)(2N^4+23N^2+191)/(2^3 \cdot 3^3 \cdot 5 \cdot 7)$

$F(4,N) = N^4(N-1)(N+1)(3N^6+43N^4+337N^2+2497)/(2^4 \cdot 3^4 \cdot 5^2 \cdot 7)$

$F(5,N) = N^5(N-1)(N+1)(4N^8 + 70N^6 + 642N^4 + 4250N^2 + 29594)/(2^6 \cdot 3^5 \cdot 5 \cdot 7 \cdot 11)$

and just in case you think you've spotted a pattern:

$F(6,N) = N^6(N-1)(N+1)(1382N^{10} + 28682N^8 + 307961N^6 + 2295661N^4+13803157N^2+92427157)/k$

where $k=2^6\cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13$

Also, for all $m$,

$F(m,2) = 1$

$F(m,3) = 2^{m+1}$

$F(m,4) = 2^{m}(2\cdot2^m+1)$

$F(m,5) = 2((5+\sqrt{5})^m + (5-\sqrt{5})^m)$

$F(m,6) = (3^m + 1)(2^{2m+1}+1)-1$

$F(m,7) = $?

$F(m,8) = 2^{2m}(2^{m+1}(2+\sqrt{2})^m+2^{m+1}+2^{m+1}(2-\sqrt{2})^m+1)$

[each of these can be shown from the original definition of $F$ using known values of $\text{cosec}(\pi/N)$]

Answer 1

I'll be using the generating function as pointed by Ira Gessel. Check the reference to Stanley's very interesting work, in particular on page 491. So, let $$f_N(x):=1-\frac{N\,x\,\cot(N\,\sin^{-1}x)}{\sqrt{1-x^2}}=\sum_{m=1}^{\infty}(2/N)^mF(m,N)\,x^{2m}.$$ Let's work with $g_N(x):=\log(\sin(N\sin^{-1}x))$ as it is related to $f_N=1-x\frac{d g_N}{dx}$. If $y=\cos\theta$ then Tchebychev polynomials of the $2^{nd}$-kind can be given by $U_{N-1}(\cos\theta)=\frac{\sin(N\,\theta)}{\sin\theta}$ and more explicitly $$U_{N-1}(y)=\sum_{k=0}^{\lfloor\frac{N-1}2\rfloor}\binom{N}{2k+1}\,(y^2-1)^k\,y^{N-1-2k}.$$ Tayloring for our purpose, $\sin(N\sin^{-1}x)=\sin(\sin^{-1}x)\cdot U_{N-1}(\sqrt{1-x^2})$. Hence, $$\sin(N\sin^{-1}x)=x\cdot \sum_{k=0}^{\lfloor\frac{N-1}2\rfloor}(-1)^k\binom{N}{2k+1}\,x^{2k}\,(1-x^2)^{\frac{N-1}2-k}.$$ Case 1: $N=2n+1$ is odd. Denote, for a short hand, $$P_n(x):=\sum_{k=0}^n(-1)^k\binom{2n+1}{2k+1}\,x^{2k}\,(1-x^2)^{n-k}$$ which has constant term $P_n(0)=N$. Further designate $Q_n(x):=P_n(x)-P_n(0)$. Then, we have \begin{align*} f_N&=1-x\frac{d}{dx}\left[\log x+\log\left(N+Q_n(x)\right)\right] =-x\frac{d}{dx}\left[\log N+\log\left(1+\frac1NQ_n(x)\right)\right] \\ &=x\frac{d}{dx}\left[-\log\left(1+\frac1NQ_n(x)\right)\right] =x\frac{d}{dx}\left[\sum_{m=1}^{\infty}\frac1m\cdot\left(\frac1{N}Q_n(x)\right)^m\right] \\ &=\sum_{m=1}^{\infty}\frac1{N^m}\cdot Q_n(x)^{m-1}\cdot x\frac{d}{dx}Q_n(x). \end{align*} At this stage, observe that the coefficients of $Q_n(x)$ are divisible by $2$, hence $Q_n(x)^{m-1}$ has a factor of $2^{m-1}$; while $\frac{dQ_n}{dx}$ contributes a factor of $2$ from its coefficients. All together, the polynomial $\frac{N^m}{2^m}\left(\frac{Q_n(x)^{m-1}}{N^m}\cdot x\frac{d}{dx}Q_n(x)\right)$ has integer coefficients. That means the coefficient of $\frac{N^m}{2^m}\left(\frac{2^m}{N^m}F(m,N)\right)$ are indeed integers, as required.

Case 2: $N=2n$ even. The argument is analogous (there is an extra factor which is easily handled).

Answer 2

It turns out rewriting the function in terms of Bernouilli numbers etc, while useful in generating specific values of $F(m,N)$, is not an easy path to proving they are integers. The following proof is reasonably elementary (knowledge of complex $n$th roots of $1$, no calculus) and is drawn from D. Zagier, Elementary Aspects of the Verlinde Formula and of the Harder-Narasimhan-Atiyah-Bott Formula, Israel Math. Conf. Proc. 9 (1996) - see Theorem 1 part (vii). (I've given more detail on some of the steps, and structured it to focus on the proof of integral value).

PS - I'm an actuary, with no academic credentials beyond a Maths BA from 30 years ago, but I've enjoyed reading your comments and finding a solution to a tweet that I read 1 year ago (almost to the day!) and which has bugged me ever since.

APPROACH

Fix N.

(1) We will define a function $f(g,\mathbf{x})$ on integer $g \geq 0$ and $\mathbf{x}=(x_1,x_2,...,x_{N-1})$ for integer $x_i \geq 0$.

Then, defining $\mathbf{e_1}=(1,0,0,...0), \mathbf{e_2}=(0,1,0,...,0), ... \mathbf{e_{N-1}}=(0,0,...0,1)$

we will show

(2) $f(m+1,\mathbf{0}) = F(m,N)$

(3) $f(0,\mathbf{0})$ is an integer

(4) $f(0,\mathbf{e_t})$ is an integer

(5) $f(0,\mathbf{e_t+e_u})$ is an integer

(6) $f(0,\mathbf{e_t+e_u+e_v})$ is an integer

(7) $\displaystyle \sum_{t} f(0,\mathbf{x+e_t})f(0,\mathbf{y+e_t})=f(0,\mathbf{x+y})$ for all $\mathbf{x,y}$

(8) using (3) - (7), $f(0,\mathbf{x})$ is an integer for all $\mathbf{x}$

(9) $\displaystyle \sum_{t} f(g,\mathbf{x+2e_t})= f(g+1,\mathbf{x})$ for all $g$

(10) using (8) - (9), $f(g,\mathbf{x})$ is an integer for all $g$ and $\mathbf{x}$

and so conclude from (2) and (10) that $F(m,N)$ is an integer.

All summations are over $\{1,2,...,N-1\}$.

STEPS (1), (2)

(1)

Let

$s(z)=\displaystyle \frac{-2N}{(z - z^{-1})^2}$

$r_a(z) =\displaystyle \frac{z^{a} - z^{-a}}{z - z^{-1}}$

$r(z,\mathbf{x}) = \displaystyle\prod_a r_a(z)^{x_a}$

$\omega = e^{i \pi / N}$

$f(g,\mathbf{x})=\displaystyle \sum_{j=1}^{N-1} s(\omega^j)^{g-1}r(\omega^j,\mathbf{x})$

(2)

$s(\omega^j) = \displaystyle\frac{-2N}{(2i\sin(\pi j / N))^2}=\frac{N}{2\sin^2(\pi j / N)}\quad$ and $\quad r(\omega^j,\mathbf{0})=1\quad$ so

$f(m+1,\mathbf{0})=(\frac{N}{2})^m\sum\frac{1}{\sin^{2m}(\pi j/N)} = F(m,N)$

USEFUL RESULTS

For the next steps, the following will be useful.

Let $T_k=\displaystyle \sum_j \omega^{jk} + \omega^{-jk}$ and $U_k=\displaystyle\sum_j \frac{\omega^{jk}-\omega^{-jk}}{\omega^j-\omega^{-j}}$

Then (derivation in footnote):

$\begin{cases} T_k = 2(N-1), & U_k = 0 & \text{if $k$ is multiple of $2N$} \\ T_k = -2, & U_k = 0 & \text{if $k$ even, and not a multiple of $2N$}\\ T_k = 0, & U_k = N-k+2Nq & \text{if $k$ odd (some integer $q$)}\\ \end{cases}$

STEPS (3) - (6)

(3)

$f(0,\mathbf{0}) = \displaystyle \sum_j \frac{(\omega^j - \omega^{-j})^2}{-2N}=\frac{-1}{2N}\sum (\omega^{2j}+\omega^{-2j}-2)=\frac{-1}{2N}(T_2 - 2(N-1)) = 1$

(4)

$r(\omega^j, \mathbf{e_t})= \displaystyle \frac{\omega^{jt} - \omega^{-jt}}{\omega^j - \omega^{-j}}$,

so $r(\omega^j,\mathbf{e_1})=1$, so $f(0,\mathbf{e_1})=1$

For $t>1$,

$f(0,\mathbf{e_t})=\displaystyle -\frac{1}{2N} \sum_j (\omega^j-\omega^{-j})(\omega^{jt} - \omega^{-jt})=-\frac{1}{2N}(T_{t+1}-T_{t-1})=0$

(5)

$r(\omega^j,\mathbf{e_t+e_u})=\displaystyle \frac{\omega^{jt} - \omega^{-jt}}{\omega^j - \omega^{-j}}\frac{\omega^{ju} - \omega^{-ju}}{\omega^j - \omega^{-j}}$ so

$f(0,\mathbf{e_t+e_u})=-\displaystyle\frac{1}{2N}\sum_j (\omega^{j(t+u)}+\omega^{-j(t+u)}-\omega^{j(u-t)}-\omega^{-j(u-t)})$

$=-\frac{1}{2N}(T_{t+u}-T_{t-u})$ from which

$f(0,\mathbf{e_t+e_u})= \begin{cases} 1 & \text{if }t = u \\ 0 & \text{if }t \neq u \end{cases}$

(6)

$f(0,\mathbf{e_t+e_u+e_v})=\displaystyle -\frac{1}{2N}\sum_j\frac{(\omega^{jt} - \omega^{-jt})(\omega^{ju} - \omega^{-ju})(\omega^{jv} - \omega^{-jv})}{\omega^j - \omega^{-j}}$

$=-\frac{1}{2N}(U_{t+u+v}-U_{t+u-v}+U_{t-u-v}-U_{t-u+v})$

If $t+u+v$ etc are even, then this is $0$.

If $t+u+v$ etc are odd, then (since $U_k = N - k + \text{multiple of $2N$}$)

$U_{t+u+v}-U_{t+u-v}+U_{t-u-v}-U_{t-u+v}$

$=(t+u+v)-(t+u-v)+(t-u-v)-(t-u+v)+2Nq = 2Nq$ for integer $q$.

So $f(0,\mathbf{e_t+e_u+e_v})$ is an integer.

ANOTHER USEFUL RESULT

Let $a_{jk} =\displaystyle \sum_t r_t(\omega^j)r_t(\omega^k)=\frac{\sum (\omega^{jt} - \omega^{-jt})(\omega^{kt}-\omega^{-kt})}{(\omega^j-\omega^{-j})(\omega^k-\omega^{-k})}$

Along similar lines to (4), this is $0$ except

$a_{jj}=\displaystyle \frac{1}{(\omega^j - \omega^{-j})^2}(T_{2j}-2(N-1))=\frac{-2N}{(\omega^j - \omega^{-j})^2}$

So $a_{jk}= \begin{cases} s(\omega^j) & \text{if }j=k\\ 0 & \text{otherwise} \end{cases}$

STEPS (7) - (10)

(7)

$\displaystyle \sum_{t} f(0,\mathbf{x+e_t})f(0,\mathbf{y+e_t})$

$=\displaystyle \sum_t \left[ \sum_j s(\omega^j)^{-1}\prod_a r_a(\omega^j)^{x_a}\cdot r_t(\omega^j) \sum_k s(\omega^k)^{-1}\prod_b r_b(\omega^k)^{y_b}\cdot r_t(\omega^k) \right]$

$=\displaystyle \sum_j \sum_k s(\omega^j)^{-1}s(\omega^k)^{-1}\prod_a r_a(\omega^j)^{x_a}\prod_b r_b(\omega^k)^{y_b}\sum_t r_t(\omega^j) r_t(\omega^k)$

$=\displaystyle\sum_j s(\omega^j)^{-1} s(\omega^j)^{-1}\prod_a r_a(\omega^j)^{x_a+y_b}s(\omega^j)$ using the previous result for $a_{jk}$

$=f(0,\mathbf{x+y})$

(8)

Let $A_n = \{\mathbf{x}:x_i \geq 0, x_1+x_2+...+x_{N-1} = n\}$

We prove by induction on $n$, that $f(0,\mathbf{x})$ is an integer for $\mathbf{x} \in A_n$

The results from (3) - (6) show this is the case for $n \leq 3$.

So pick $n \geq 4$ and assume that the function is an integer for $A_m$ for all $m < n$.

$\mathbf{x} \in A_n \implies$ for some $a$ and $b$, $\mathbf{x-e_a-e_b} \in A_{n-2}$

so applying (7), $\displaystyle \sum_t f(0,\mathbf{x-e_a-e_b+e_t})f(0,\mathbf{e_a+e_b+e_t})=f(0,\mathbf{x})$

So $f(0,\mathbf{x})$ is an integer, because $\mathbf{x-e_a-e_b+e_t} \in A_{n-1}$ and $\mathbf{e_a+e_b+e_t} \in A_3$ so these give integer values by induction assumption.

(9)

$\displaystyle \sum_{t} f(g,\mathbf{x+2e_t})=\sum_t \sum_j s(\omega^j)^{g-1}r(\omega^j,\mathbf{x})r_t(\omega^j)^2$

$=\displaystyle \sum_j s(\omega^j)^{g-1}r(\omega^j,\mathbf{x})\sum_t r_t(\omega^j)r_t(\omega^j) = \sum_j s(\omega^j)^gr(\omega^j,\mathbf{x})$ using the $a_{jj}$ result

so $\displaystyle \sum_{t} f(g,\mathbf{x+2e_t})=f(g+1,\mathbf{x})$

(10)

It is a straightforward induction on $g$ using (8) and (9), to deduce that $f(g,\mathbf{x})$ is an integer for all $g$.

This completes the proof.

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FOOTNOTE

Note that $\omega^{Nk} = (-1)^k$ and $\omega^{-1} = \omega^{2N-1}$

If $k$ is a multiple of $2N$ then $\omega^k = \omega^{-k}=1$, so $T_k = 2(N-1)$, $U_k=0$

Otherwise, summing as geometric progressions,

$T_k= \displaystyle \frac{\omega^k-\omega^{Nk}+\omega^{(N+1)k}-1}{1-\omega^k}$ which simplifies to

$-2$ if $k$ is even and $0$ if $k$ is odd.

It is straightforward to show that

$\displaystyle\frac{\omega^{jk}-\omega^{-jk}}{\omega^j-\omega^{-j}}=\omega^{j(k-1)}+\omega^{-j(k-1)}+\omega^{j(k-3)}+\omega^{-j(k-3)}+...+\omega^{jp}+\omega^{-jp}+a$

where

$p = 2$ and $a=1$ if $k$ is odd,

$p=1$ and $a=0$ if $k$ is even.

Thus $U_k=T_{k-1} + T_{k-3} + ... T_p+(N-1)a$

If $k$ is even then $U_k=0$

If $k$ is odd, then $T_{k-1} + ... + T_p$ contains $(k-1)/2$ instances of $T_{\text{even}}$ which are either $-2$ or $-2+2N$, so $U_k = -(k-1) + 2Nq + (N-1)$ where $q$ counts the occurrences of multiples of $2N$ in $\{2, 4, ... k-1\}$.

So, if $k$ is odd, $U_k = N-k+2Nq$ for some integer $q$.