This answer is free for anyone to use.

Let's start from your formula _{(I assume your notation was in error using $a_N$ instead of $a_k$)}

$$ \prod_{k=1}^{N}\left(3+\frac{1}{a_k}\right)=2^S \tag 1 $$ So $a_1 (=a_\min )$ must be smaller than some "mean"-value $\alpha(N)$ and some $a_\max$ value must be larger than that $\alpha(N)$. Let's rewrite $$ 2^S = \prod_{k=1}^{N}\left(3+\frac{1}{\alpha}\right) \qquad \qquad \text{and note the rhs is also} = \left(3+\frac1{\alpha} \right)^N $$ Then $$ 2^S / 3^N = \left(1+\frac1{3\alpha} \right)^N \tag 2 $$ and then $$ S \cdot l2 - N\cdot l3 = N \cdot \log\left(1+\frac1{3\alpha} \right) \tag 3 \\ \small \text{ where } l2 := \log(2) \text{ and } l3:= \log(3)$$

The "G.Rhin-bound" (used by J. Simons 2003)

We know, that the lhs has a lower bound, one given by S. Ellison (used for instance in T.Tao's blog), one by G. Rhin (used by R. Steiner/J. Simons) so for instance the Rhin-bound means $$ {1 \over 457 \cdot N^{13.3} } \lt S \cdot l2 - N\cdot l3 \tag {4.a} $$ and from this $$ {1 \over 457 \cdot N^{13.3} } \lt N \cdot \log\left(1+\frac1{3\alpha} \right) \tag {4.b} $$ Doing this further $$ \begin{array} {llll} \qquad &{1 \over 457 \cdot N^{13.3} } &\lt & N \cdot \left(\frac1{3 \alpha}-\frac1{2 \cdot 3^2 \cdot \alpha^2 } + \cdots \right) \\ &{1 \over 457 \cdot N^{13.3} } &\lt& N \cdot \frac1{3 \alpha} \\ &\alpha &\lt& \frac{N}3 \cdot 457 \cdot N^{13.3} \end{array} \tag {4.c} $$

$$ \begin{array} {llll} (a_1 \lt) & \alpha & \lt & 153 \cdot N^{14.3} \qquad \qquad . \end{array} \tag {4.d}$$

My (2009) "N log N"-conjecture
I've conjectured, based on empirical evidences up to $N=10^{600 \, 000}$, that $$ {1 \over 10 N \ln N } \lt S \cdot l2 - N\cdot l3 {\qquad\text{ N Log N-conjecture}} \tag {5.a} $$ Assuming that this is indeed true for all larger $N$, then starting from $$ S \cdot l2 - N\cdot l3 = N \cdot \log\left(1+\frac1{3\alpha} \right) \\ \small \text{ where } l2 := \log(2) \text{ and } l3:= \log(3)$$ we have as well $$ {1 \over 10 N \ln N } \lt N \cdot \log\left(1+\frac1{3\alpha} \right) \\ {1 \over 10 N \ln N } \lt N \cdot \frac1{3\alpha} \tag {5.b} $$ and then $$ (a_1 \lt ) \qquad \alpha \lt \frac1{3} \cdot 10 N^2 \ln N \\ \tag {5.c} $$ which gives a much sharper upper bound for $\alpha$ and thus for the smallest element in some assumed cycle $a_1$

This answer is free for anyone to use.

Until I find a way to avoid shorteners and keep the list in my profile.

• A short proof of Stirling's inequality through creative telescoping ^[1]
• A Ramanujan sum ^[2]
• The Russian integral ^[3]
• Fibonacci number that ends with 2014 zeroes ^[4]
• Fibonacci infinite sum resulting in $\pi$ ^[5]
• An analogue of Hensel lifting for Fibonacci numbers ^[6]
• An interesting identity involving odd powers of $\pi$ ^[7]
• The sum of $\frac{1}{n\sinh(\pi n\sqrt{3})}$ on odd numbers ^[8]
• How to prove that $\sum_{k\geq 1}\frac{\zeta(2k)}{2^{2k-1}}=1$ ^[9]
• How to distinguish walking on a sphere or on a torus ^[10]
• Is $\sum_{n\geq 1}\frac{\left|\sin n\right|^n}{n}$ convergent? (YES it is) ^[11]
• Sum of squares of harmonic numbers ^[12]
• A closed form for $\phantom{}_4 F_3\left(1,1,1,\frac{3}{2};\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)$ ^[13]
• On the largest root of the Hermite polynomial $H_n$ ^[14]
• Fourier transform of squared Gaussian Hermite polynomial ^[15]
• General formula for the 1;5;19;65;211 sequence ^[16]
• A NASTY integral of a rational function ^[17]
• Integral $\int_{0}^{+\infty}\frac{\text{arccot}(\sqrt{x}-2\sqrt{x+1})}{x+1}dx$ ^[18]
• Integral $\int_{0}^{\pi}\arctan^2\left(\frac{\sin x}{2+\cos x}\right)dx$ ^[19]
• Different methods to compute $\zeta(2)=\sum_{k\geq 1}\frac{1}{k^2}$ (aka Basel problem) ^[20]
• On the increasing nature of $x^{x^{x^\ldots}}$ on $\left[1,e^{1/e}\right]$ ^[21]
• A closed form for $\sum_{k\geq 0}\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)$ (aka the Taylor series of the squared arcsine) ^[22]
• An integer-valued power sum of secants ^[23]

[1] https://math.stackexchange.com/questions/94722
[2] https://math.stackexchange.com/questions/2442091
[3] https://math.stackexchange.com/questions/2529614
[4] https://math.stackexchange.com/questions/872071
[5] https://math.stackexchange.com/questions/1424554
[6] https://math.stackexchange.com/questions/872955
[7] https://math.stackexchange.com/questions/850442
[8] https://math.stackexchange.com/questions/1330740
[9] https://math.stackexchange.com/questions/1080000
[10] https://math.stackexchange.com/questions/854380
[11] https://math.stackexchange.com/questions/823816
[12] https://math.stackexchange.com/questions/933656
[13] https://math.stackexchange.com/questions/2123298
[14] https://math.stackexchange.com/questions/2525094
[15] https://math.stackexchange.com/questions/2626919
[16] https://math.stackexchange.com/questions/1790642
[17] https://math.stackexchange.com/questions/266181
[18] https://math.stackexchange.com/questions/557439
[19] https://math.stackexchange.com/questions/1593334
[20] https://math.stackexchange.com/questions/8337
[21] https://math.stackexchange.com/questions/1335096
[22] https://math.stackexchange.com/questions/878477
[23] https://math.stackexchange.com/questions/5090904

Starting from $t^2-(xy+x+y)^3+x^2y^2$ over an algebraically closed field of characteristic 3, first look at the degree-three portion, $-x^3-y^3$, and apply $x\mapsto -x-y$ to send the degree-three portion to $x^3$ and our defining polynomial to $$ t^2 + x^3 + y^4 + x^2 y^2 - x y^3 + x^3 y^3 + y^6. $$ Now we're very close on the degree-four portion except for $-xy^3$. To get rid of that, apply $y\mapsto y+x$ to get $$ t^2 + x^3 + y^4 + x^2 y^2 + x^4 - x^6 + y^6. $$

Getting rid of $x^4$ is trickier. There are two options:

1. Try and get a $-x^4$ term from somewhere
2. Try and get $t^2x+y^4x+x^3y^2-x^7+xy^6$ so we can get to $(1+x)(t^2+x^3+y^4+x^2y^2+\cdots)$.

$$ 2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$ $$ 3^n-2^n+2^n<2^{\lceil n \log_23\rceil}-3^n+3^n-2^n<2(3^n-2^n)\\ 3^n<2^{\lceil n \log_23\rceil}-2^n<2(3^n-2^n)\\ 3^n<2^n(2^{S-n}-1)<2(3^n-2^n)\\ (3/2)^n<2^{S-n}-1< 2 (3/2)^n-2\\ (3/2)^n+1<2^{S-n}< 2\cdot(3/2)^n-1\\ $$

This answer is free to use by everyone.

[Work in progress (08/08/2025)]

Context

For $n\in\mathbb{N}^{+}$, I define $$\bbox[15px,#FFECEC,border:2px groove #E74C3C]{ \delta_{z|n}=\frac{1}{n}\sum_{\omega:\,\omega^n=1}\omega^z}$$ where $\omega$ are the roots of unity ^[1] where $\text{arg}(\omega)\in(-\pi,\pi]$.

In particular:

• $\displaystyle \delta_{z|1}=1$
• $\displaystyle \delta_{z|2}=\frac{1}{2}\left(1+e^{\pi i z}\right)$
• $\displaystyle \delta_{z|3}=\frac{1}{3}\left(1+2\cos\left(\frac{2\pi}{3}z\right)\right)$
• $\displaystyle \delta_{z|4}=\frac{1}{4}\left(1+2\cos\left(\frac{\pi}{2}z\right)+e^{\pi i z}\right)$
• $\displaystyle \delta_{z|5}=\frac{1}{5}\left(1+2\cos\left(\frac{2\pi}{5}z\right)+2\cos\left(\frac{4\pi}{5}z\right)\right)$
• etc...

Properties

This function has a lot of properties:

$1)$ Explicit formula:
$$\bbox[15px,#FFF4E0,border:2px groove #F39C12]{\displaystyle \delta_{z|n}=\frac{1}{n}\left(1+2\sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}\cos\left(\frac{2k\pi}{n}z\right)+\text{mod}(n-1,2)e^{\pi iz}\right)\quad\Rightarrow\quad \delta_{z|2n+1}\in\mathbb{R}}$$

$2)$ Specific value in $j\in\mathbb{Z}$
$$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{\displaystyle \delta_{j|n}=\begin{cases}1&\text{if }j\text{ is a multiple of }n\\ 0&\text{otherwise}\end{cases}}$$

$3)$ Conjugate symmetric ^[2]
$$\bbox[15px,#F5EEFB,border:2px groove #9B59B6]{\displaystyle \delta_{-z^{*}|n}^{*}=\delta_{z|n}}$$

$4)$ Periodicity:
$$\bbox[15px,#E0EFFF,border:2px groove #0058B1]{\displaystyle \delta_{z-n|n}=\delta_{z|n}}$$

$5)$ Link with the sinc function ^[3]:
$$\bbox[15px,#FFF4E0,border:2px groove #F39C12]{\displaystyle \lim_{n\to\infty}\delta_{z|n}=\text{sinc}(z)=\begin{cases}\frac{\sin(\pi z)}{\pi z}&z\neq 0\\1&z=0\end{cases}}$$

$6)$ Closed under circular convolution ^[4] (discrete and continue) $$\bbox[15px,#F5EEFB,border:2px groove #9B59B6]{\displaystyle (\delta_{m}\ast\delta_n)(z)=\begin{cases} \displaystyle\sum_{k=1}^{\text{lcm}(m,n)}\delta_{k|m}\delta_{z-k|n}\\ \displaystyle\int_0^{\text{lcm}(m,n)}\delta_{t|m}\delta_{z-t|n}\mathrm{d}t\end{cases}=\delta_{z|\gcd(m,n)}}$$

In particular,
$6.1)$ If $m=1$, convex combination (discrete and continue):
$$\bbox[15px,#E9F7EF,border:2px groove #27AE60]{\displaystyle \sum_{k=1}^{n}\delta_{z-k|n}=1\qquad\qquad \int_0^{n}\delta_{t|n}\mathrm{d}t=1}$$ $6.2)$ If $m=n$, idempotent function under discrete and continue circular convolution:
$$\bbox[15px,#E0EFFF,border:2px groove #0058B1]{\displaystyle \sum_{k=1}^n \delta_{k|n}\delta_{z-k|n}=\delta_{z|n}\qquad\qquad\int_0^n \delta_{t|n}\delta_{z-t|n}\mathrm{d}t=\delta_{z|n}}$$ $6.3)$ If $m=n$ and $z\in \mathbb{Z}$, I get that has orthogonal translations in $\ell^2([0,n])$ and $L^2([0,n])$:
$$\bbox[15px,#E0EFFF,border:2px groove #0058B1]{\displaystyle\sum_{k=1}^{n}\delta_{k|n}\delta_{k-j|n}^{*}=\delta_{j|n}\qquad\qquad\int_{0}^{n}\delta_{t|n}\delta_{t-j|n}^{*}\mathrm{d}t=\delta_{j|n}}$$ $6.4)$ If $m$ and $n$ are coprime ^[5] ($\gcd(m,n)=1$): $$\bbox[15px,#F5EEFB,border:2px groove #9B59B6]{\displaystyle \sum_{k=1}^{mn}\delta_{k|n}\delta_{z-k|m}=1\qquad\qquad\int_0^{mn}\delta_{t|n}\delta_{z-t|m}\mathrm{d}t=1}$$ $6.5)$ For properties 3 and 4 it is closed under circular cross-correlation ^[6] (discrete and continue) $$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{\displaystyle \begin{array}{c} ( \delta_{|n}\star \delta_{|m} )(z)=\begin{cases} \displaystyle\sum_{k=1}^{\text{lcm}(m,n)} \delta_{k|n}^{*}\delta_{z+k|m}\\ \displaystyle\int_0^{\text{lcm}(m,n)} \delta_{t|n}^{*}\delta_{z+t|m}\mathrm{d}t\end{cases}=\delta_{z|\text{gcd}(m,n)} \end{array}} $$ In particular,
$6.6)$ If $m=n$ idempotent function under discrete and continue cross-correlation (similar to $6.2$)
$6.7$) For $z=0$: inner product in $\ell^2([0,\text{lcm}(m,n)])$ and $L^2([0,\text{lcm}(m,n)])$: $$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{\displaystyle \begin{array}{c} \displaystyle\left\langle \delta_{|n}, \delta_{|m} \right\rangle_{\ell^2([0,\text{lcm}(m,n)])}=\sum_{k=1}^{\text{lcm}(m,n)} \delta_{k|n}\delta_{k|m}^{*}=1\\ \displaystyle\left\langle \delta_{|n},\delta_{|m}\right\rangle_{L^2([0,\text{lcm}(m,n)])}=\int_0^{\text{lcm}(m,n)} \delta_{t|n}\delta_{t|m}^{*}\mathrm{d}t=1 \end{array}} $$

$7)$ [From the comments] it can be seen as shifted inverse DFT ^[7] of $(1,1,1,...)$ $$\bbox[15px,#FFF4E0,border:2px groove #F39C12]{\delta_{z|n}=\frac{1}{n}\sum_{k=\left\lceil\frac{1-n}{2}\right\rceil}^{\left\lfloor\frac{n}{2}\right\rfloor}e^{-\frac{2\pi ik}{n}x}=\frac{1}{n}\sum_{k=1}^{n}\exp\left(i\arg\left(e^{\frac{2k\pi i}{n}}\right)x\right)}$$

$\color{red}{\text{[Just of notation]}}$ in analogy with the continuous Fourier transform and the 2th property, I think it might make sense to indicate $\delta_{z|n}$ as a sort of analytical continuation of a periodic Kroneker $\delta$ with period $n$

8. Multidimensional extension:
   I define $$\bbox[15px,#FFECEC,border:2px groove #E74C3C]{\displaystyle\delta_{z_1,...z_d|n}=\frac{1}{n^d}\sum_{\omega:\,\omega^n=1}\omega^{z_1+...+z_d}}$$ The properties on individual arguments continue to apply, in particular I define $\mathbf{z}=(z_1,...,z_d)$, $\mathbf{k}=(k_1,...,k_d)$ and $\mathbf{t}=(t_1,...,t_d)$ $$\bbox[15px,#E0FFFF,border:2px groove #1ABC9C]{(\delta_{|n}\ast\delta_{|m})(\mathbf{z})=\begin{cases}\displaystyle\sum_{k_1=1}^{\text{lcm}(m,n)}\cdots\sum_{k_d=1}^{\text{lcm}(m,n)}\delta_{\mathbf{k}|m}\delta_{\mathbf{z}-\mathbf{k}|n}\\ \displaystyle\int_{[0,\text{lcm}(m,n)]^d}\delta_{\mathbf{t}|m}\delta_{\mathbf{z}-\mathbf{t}|n}\mathrm{d}\mathbf{t}\end{cases}=\delta_{\mathbf{z}|\gcd(m,n)}}$$

$$\delta_{z_1,...,z_d|n}=n^{d-1}\delta_{z_1+...+z_d|n}$$ $$\delta_{z_1^{*},...,z_d^{*}|n}^{*}=\delta_{-z_1,...,-z_d|n}$$ $9)$ Work in progress...

Question

Has this function a name? Or is it connected to some known function?

This looks like the Dirichlet kernel ^[8] but it is not the same.

[1] https://en.wikipedia.org/wiki/Root_of_unity
[2] https://en.wikipedia.org/wiki/Even_and_odd_functions#Complex-valued_functions
[3] https://en.wikipedia.org/wiki/Sinc_function
[4] https://en.wikipedia.org/wiki/Circular_convolution
[5] https://en.wikipedia.org/wiki/Coprime_integers
[6] https://Cross-correlation
[7] https://en.wikipedia.org/wiki/Discrete_Fourier_transform
[8] https://en.wikipedia.org/wiki/Dirichlet_kernel

This answer is free for anyone to use.

The beginnings of an answer to Sum and difference of integers on a blackboard ^[1].

I believe the answer is that $N=n-1$ if $n\equiv 0,3\pmod 4$ and $N=n-2$ if $n\equiv 1,2\pmod 4$. I can show that $N\geq n-2$ always, and that $N$ is at most this conjectured value.

---

We first show the upper bound. Since the sum of the numbers is nondecreasing, we always $N\geq n-1$. Also, note that the parity of sum of the numbers on the board never changes. Therefore, if $n\equiv 1,2\pmod 4$, there is always at least one odd number on the board. However, if we are to end up with only one nonzero number on the board, it must be even: to get to $m,0,\ldots,0$, we must have gone from $m/2,m/2,0,\ldots,0$, and so $m$ must be even. So, if $n\equiv 1,2\pmod 4$, there will always be at least two nonzero numbers on the board.

---

We now show the lower bound, providing a procedure to get at least $n-2$ zeros.

Lemma 1. If we have $x,x,0$ on the board, we can, in some finite numbers of moves, replace these numbers with $2^kx,0,0$ for any $k\geq 1$.

Proof. For $k=1$, we can simply operate on $(x,x)$. So, it suffices to show that we can turn $x,x,0$ into $2x,2x,0$. Indeed, we do $$(x,x,0)\to (x,x,2x)\to (2x,0,2x).\square$$

Lemma 2. Suppose the board contains a zero and two ones, all nonzero numbers on the board are all powers of two, and there are an even number of ones. Then we can make it so there is only one nonzero number on the board.

Proof. Set aside two ones and all the zeros, and call the remaining numbers "living." First, while the positive living numbers are not all distinct, replace $x,x$ with $2x,0$. This increases the number of zeros, and so must terminate. This also does not change the parity of the number of ones; in particular, at the end of this procedure, there are no living ones. So, the nonzero numbers on the board are $$1,1,2^{a_1},2^{a_2},\ldots,2^{a_k}$$ for some $0<a_1<\cdots<a_k$. Now, replace $(1,1,0)$ with $(2^{a_1},0,0)$ by Lemma 1, so that the nonzero numbers on the board are $$2^{a_1},2^{a_1},2^{a_2},\ldots,2^{a_k}.$$ Now, replace $(2^{a_1},2^{a_1},0)$ with $(2^{a_2},0,0)$ by Lemma 1. We can repeat this process until we are left with two copies of $2^{a_k}$ and some zeros, at which point we can replace these two copies with $(2^{a_k+1},0)$. $\square$

Lemma 3. Suppose there are $n$ numbers on the board. If every number on the board is a power of two (in particular, there are no zeros), and the largest number is at most $2^{n-2}$, then there is a sequence of moves to produce $n-2$ zeros.

Proof. Let $a$ be the smallest power of two on the board, and let $b$ be a power of two which appears twice, guaranteed to exist because there are at most $n-1$ distinct numbers on the board. If $a=b$, then we can produce $n-1$ zeros by Lemma 2 (scaling up all numbers by $a$ in Lemma 2 changes nothing). Otherwise, operate on $(b,b)$ to get $(0,2b)$.

[1] https://math.stackexchange.com/questions/4969636/

This answer is free for anyone to use.

This post if free to use for any one.

This post is free to use for anyone.

This answer is free for anyone to use.

This answer is free for anyone to use.

This answer is free for anyone to use.

For $n>0$:$y^{(n)}=cy^n\implies y=\sum_{j=0}^\infty a_jx^j=a_0+\dots +a_{n-1}x^{n-1}+\frac{a_0^nc}{n!}x^n+ \frac{c n a_0^{n-1} a_1}{(n+1)!}x^{n+1}+ \frac{c n a_0^{n-2} ((n-1) a_1^2 + 2 a_0 a_2)}{(n+2)!}x^{n+2}+ \frac{c n a_0^{n-3 } ((n-1) a_1 ((n-2) a_1^2 + 6 a_0 a_2) + 6 a_0^2 a_3)}{(n+3)!}x^{n+3}+\dots\\a_0j^{(n+1)}a_j= -\sum_{k=1}^{j-1}k^{(n)} (k-n - n (j-k))a_ka_{j-k},a_n=\frac{a_0^nc}{n!}$$ For $-n=m>0$: $$\int\dots\int ydx\dots dx=cy^{-m}\implies y=\sum_{j=0}^\infty a_j x^j\\$$

$n=-1$ case ^[1]

$n=-2$ case ^[2]

---

$\def \F{\text F}$ To derive a power series for the inverse of $f(z)=\,_2\F_1(a,b,c,z)$, start with the hypergeometric differential equation ^[3], substitute $y\to y’$, and transform it into one for $\frac c{ab}(y-1)$ apply the inverse substitution ^[4] $y\to z,y’\to\frac1{y’},Y=\int_0^zy(t)dt\to yz-Y$

$$z(1-z)y’’+(c-(a+b+1) z)y’-a b y=0,y(0)=1,y’(0)=\frac{ab}c\\\implies v(1-v) y’ +(c-1-(a+b-1)v)y-(a-1) (b-1) Y-c v=0,y(0)=0,y’(0)=1\\\implies ((a b v+c) y-(a-1)(b-1) Y-(c-1)v)y’-y+y^2=0,y(0)=0,y’(0)=1$$

One now can find $y=\sum\limits_{n=1}^\infty a_n v^n,v=\frac c{ab}(z-1)$. Substituting it and gathering powers of $v$ gives:

$$_2\F_1^{-1}(a,b,c,z)=\sum_{n=1}^\infty a_n\left(\frac c{ab}(z-1)\right)^n,(c-1+n)a_n=\sum_{k=1}^{n-1}\left(\left(\frac{(a - 1) (b - 1)}{k+1}-a b\right) (n-k)-1\right) a_{n-k} a_k-\sum_{k=2}^{n-1}ck a_{n-k+1} a_k $$

---

To derive a Puisex ^[5] like series for $y=f^{-1}(z)$, start with its differential equation $(1)$, notice $y\sim \frac cb\left(1-z^{-\frac1a}\right),|z|\to0$, and let $z=(1-\frac bcv)^{-a}[]$:

$$y(y-1)(c-bv)y’’-y((c-bv)y’((a+b+1)y-c)+(a+1)by(y-1)+(c-bv)^2y’^2)=0$$

to get a series $y=\sum\limits_{n=1}^\infty a_nv^n$ in $v=\frac cb \left(1-z^{-\frac1a}\right)$. Substituting it and finding the recurrence relation via the same process gives:

$$\,_2{\F_1}^{-1}(a,b,c,z)=\sum_{n=1}^\infty a_n\left(\frac cb\left(1-z^{-\frac1a}\right)\right)^n\\\\-cn(c+n-1)a_{n+1}=\sum_{m=1}^{n-1}\sum_{j=1}^{n-m-1}a_{n-m-j}a_ma_j((a+1)bj+b^2mj(n-m-j)-b(a+b+1)mj+bj(j-1))+\sum_{m=1}^{n-1}\sum_{j=1}^{n-m}a_{n-m-j+1}a_ma_jcj((a+b+1)m-2bm(n-m-j+1)-(j-1))+\sum_{m=2}^{n-1}\sum_{j=1}^{n-m+1}c^2a_{n-m-j+2}a_ma_jmj(n-m-j+2)-b(a+1-c(n-1))a_{n-1}+\sum_{m=2}^{n-1}ma_m(c(m-1)a_{n-m+1}-b(a+m+cm-cn)a_{n-m})[]$$

---

---

$\def\dn{\operatorname{dn}}$

This goal is to understand how to expand inverses of non-elementary functions as a series. For example the Jacobi dn ^[6] Fourier cosine series from Paramanand’s blogspot ^[7]:

$$\dn(u,k)=\frac{a_0}2+\sum_{n=1}^\infty a_n\cos(2nz),a_n=\frac1\pi\int_{-\pi}^\pi\dn(u,k)e^{-2inz}dz,z=\frac{\pi u}{2K(k)}$$

using residue calculus to find $a_n=\frac{2\pi}{K(k)(q^{-n}(k)+q^n(k)}$ with the nome $q(k)$ ^[8] and complete elliptic integral of the first kind $K(k)$ ^[9]. However, for someone not knowing the residue theorem, it would be hard to derive this result.

[1] https://www.desmos.com/calculator/ntrpnhupt2
[2] https://www.desmos.com/calculator/medzvvswwo
[3] https://mathworld.wolfram.com/HypergeometricDifferentialEquation.html
[4] https://math.stackexchange.com/questions/4963131/does-y-to-x-y-to-fracy-y-3-always-convert-a-differential-equation-int
[5] https://en.wikipedia.org/wiki/Puiseux_series
[6] https://dlmf.nist.gov/22.14
[7] https://paramanands.blogspot.com/2011/02/elliptic-functions-fourier-series.html
[8] https://en.wikipedia.org/wiki/Nome_(mathematics)
[9] https://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html

$$ (2^B \cdot 2^N - (2^B \cdot n + r))\cdot k = 2^B - 1 \\ 2^B \cdot (2^N - n)\cdot k - 2^B = r\cdot k -1 \\ (2^N - n)\cdot k = {r\cdot k -1\over 2^B } +1\\ k = {r\cdot k +2^B -1\over 2^B } \\ 2^B\cdot k = {r\cdot k +2^B -1\over (2^N - n) } \\ 2^B\cdot k = {r\cdot k +2^B -1\over (2^N - n) } \\ 2^B = {r\cdot k +2^B -1\over (2^N - n)\cdot k } \\ 2^B = {r +(2^B -1)/k\over (2^N - n) } \\ $$ Here, rhs.numerator $\lt 2\cdot 2^B$. Thus rhs.denominator cannot be larger than $1$.

• • • • • ========================== $$ (2^N - n)\cdot k -1 = {r\cdot k -1\over 2^B } \\ 2^B \left((2^N - n)\cdot k -1 \right) = r\cdot k -1 \\ 2^B = { r\cdot k -1 \over (2^N - n)\cdot k -1} \\ 2^B = { r -1/k \over (2^N - n) -1/k} \\ $$

---

$$ 1\cdot k = {r\cdot k -1\over 2^B } +1 \\ k -1 = {r\cdot k -1\over 2^B } \\ (k -1) 2^B = r\cdot k -1 \\ k (2^B - r) = 2^B -1 \\ k = {2^B -1 \over 2^B - r } \\ k -1 = {2^B -1 \over 2^B - r } - 1\\ k -1 = {r -1 \over 2^B - r } \\ 2^B - r = {r -1 \over k -1 } \\ $$