Generalizing Bellard's "exotic" formula for $\pi$ to $m=11$

Question

Bellard's "exotic" pi formula found here (eq. 129) has the form,

$$a\pi+b = \sum_{n=1}^\infty \dfrac{P(n)}{{\displaystyle \binom{mn}{2n}2^{n-1}}}$$

where $(a,b,m)$ are integers and he uses order $m=7$. However, it seems there are an infinite number of such formulas for other orders $m$.

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I. Orders

Order 3:

$$\pi+6 = \sum_{n=1}^\infty \dfrac{-6+50n}{{\displaystyle \tbinom{3n}{2n}2^{n}}}\tag1$$

$$3\pi+23 = \sum_{n=1}^\infty \dfrac{7+125n^2}{{\displaystyle \tbinom{3n}{2n}2^{n}}}$$

$$91\pi+831 = \sum_{n=1}^\infty \dfrac{129+3125n^3}{{\displaystyle \tbinom{3n}{2n}2^{n}}}$$

and so on. Note: Since $\binom{mn}{pn} = \binom{mn}{(m-p)n}$, then $(1)$ is equivalent to the form by Gosper,

$$\pi = \sum_{n=0}^\infty \dfrac{-6+50n}{{\displaystyle \tbinom{3n}{n}2^{n}}}$$

Order 7:

$$740025\pi+20379280 = \sum_{n=1}^\infty \dfrac{P_1(n)}{{\displaystyle \tbinom{7n}{2n}2^{n-1}}}\tag{2}$$

$$740025\pi+19755520 = \sum_{n=1}^\infty \dfrac{P_2(n)}{{\displaystyle \tbinom{7n}{2n}2^{n-1}}}\tag{3}$$

where,

$$\small{P_1(n) = 3(10996648 - 196882274 n + 1031962795 n^2 - 2942969225 n^3 + 3125347237 n^4 - 885673181 n^5)}$$

$$\small{P_2(n) = 20202864 - 361815268 n + 1669902852 n^2 - 4185508285 n^3 + 1811392311 n^4 + 3820998353 n^5 - 2124144507 n^6.}$$

1. The eqn $(2)$ is Bellard's, $(3)$ is mine and I found there are also $P(n)$ that are $6$th, $7$th deg, and so on, with different $a\pi+b$.
2. Note that $a=740025 = 3^2\cdot5^2\cdot11\cdot13\cdot23$ factors into small primes and which is a good "test" for the next orders.

Order 11: (by yours truly)

$$7997795704284513820875\pi+186851093786889785568000= \sum_{n=1}^\infty \dfrac{P(n)}{{\displaystyle \tbinom{11n}{2n}2^{n-1}}}\tag4$$

where,

$$\small{P(n) = -1560353362660981617724800 + 50163087598613671757825520 n - 582276421453108529245554812 n^2 + 3934659571398075493770398672 n^3 - 14317202423564834332818033237 n^4 + 33962269581940193651909397387 n^5 - 43329011662268469435221715498 n^6 + 28124977321512890382308084178 n^7 - 4829379078844103835855196933 n^8 - 1529681997002493500502814877 n^9.}$$

1. There are also $P(n)$ that are $10$th, $11$th deg, and so on, with different $a\pi+b$.
2. The prime factors of $a=7997795704284513820875$ are, $$3, 5, 7, 13, 17, 19, 23, 37, 41, 43, 59, 61, 79.$$
3. For each order $m$, it seems the polynomial $P(n)$ has smallest degree $m-2$.

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III. Question

Q: I found these using Mathematica's integer relations sub-routine. It could not find similar identities for order $m=4v+1$. However, can one find identities for all $m = 4v+3$?

Answer 1

NOTE: This is not a full answer, but it is possibly a step in the right direction.

I suppose that in your question you want the assumption that $P(n)$ has rational coefficients (if not, we can easily construct an example by letting $P$ have a coefficient with $\pi$ in it for any order $m$). Moreover, we may assume that $P(n)$ has integer coefficients without loss of generality (suppose we find a suitable $P$, then we can multiply it by the LCM of all the coefficients' denominators). As such, this problem is equivalent to finding $a\in \mathbb{Z},b\in \mathbb{Z},P\in \mathbb{Z}[x]$ for a given $m$ satisfying: $$a\pi+b = \sum_{n=0}^\infty \frac{P(n)}{{\binom{mn}{2n} 2^{n}}}. \tag{$\ast$}$$ Indeed, $$\sum_{n=1}^\infty \dfrac{P(n)}{{\binom{mn}{2n}2^{n-1}}}=2\sum_{n=0}^{\infty} \frac{P(n)}{\binom{mn}{2n}2^n}-2P(0).$$ We will therefore focus our attention on $(\ast)$, as the expressions for this sum are easier.

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We now wish to compute this sum for a given $P$. We use ideas based on [1]. Note that $$\begin{align} \frac{1}{\binom{mn}{2n}}&=(mn+1)\frac{\Gamma(2n+1)\Gamma((m-2)n+1)}{\Gamma(mn+2)}\\&=(mn+1)B(2n+1,(m-2)n+1)\\&=(mn+1)\int_0^1 x^{2n}(1-x)^{(m-2)n}~dx. \end{align}$$ Hence $$\sum_{n=0}^{\infty} \frac{P(n)}{\binom{mn}{2n}2^{n}}=\int_0^1 \sum_{n=0}^{\infty} P(n)(mn+1)\left(\frac{x^{2}(1-x)^{m-2}}{2}\right)^n~dx.$$ Let $y:=\frac{x^2(1-x)^{m-2}}{2}$. Then $$\sum_{n=0}^{\infty} P(n)(mn+1)y^n=\frac{Q(y)}{(1-y)^{d+2}},$$ where $d:=\deg(P)$ and $\deg(Q)=d+1$. If one wishes, one can write down $Q$ explicitly using the closed form for $\sum_{n=0}^{\infty} n^k y^n$). Substituting back, one obtains something of the form $$\sum_{n=0}^{\infty} \frac{P(n)}{\binom{mn}{2n}2^{n}}=\int_0^1 \frac{R(x)}{(x^2(1-x)^{m-2}-2)^{d+2}}~dx, \tag{1}$$ where $\deg(R)=m(d+1)$.

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We now give a heuristic for why we expect $\pi$ to only appear when $m\equiv 3\bmod 4$. A good way for $(1)$ to give $\pi$ is to have an $\arctan(x)$ or $\arctan(x-1)$ in it's anti-derivative. Since $\int \frac{1}{x^2+1}~dx=\arctan(x)+C$ and $\int \frac{1}{(x-1)^2+1}~dx=\arctan(x-1)+C$, we want $f_m(x):=x^2(1-x)^{m-2}-2$ to have $x^2+1$ or $(x-1)^2+1$ as a factor (think about integration using partial fractions). Hence we want $f_m$ to either have a zero at $i$ or $1+i$. We see that $f_m$ doesn't have a zero at $i$ for any $m$, and we see that $f_m$ has a zero at $1+i$ if and only if $m\equiv 3\bmod 4$ ($f_{m}(1+i)=-2+2i(-i)^{m-2}$). This is why Mathematica wasn't able to find similar identities for $m\equiv 1\bmod 4$!

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We now restrict to the setting when $m=4v+3$. As such, $$\sum_{n=0}^{\infty} \frac{P(n)}{\binom{mn}{2n}2^{n}}=\int_0^1 \frac{R(x)}{(x^2(1-x)^{4v+1}-2)^{d+2}}~dx,$$ where $\deg(R)=(4v+3)(d+1)$. We have therefore related finding $P$ to finding $R$ such that the integral on the right-hand side is of the form $a\pi+b$. Taking this into account, I strongly suspect that it is possible to do it for all $v\in \mathbb{N}$ (considering that you can select $P$ to have as large of a degree as you want), but I do not have a proof of it. Any further input on this would be much appreciated.

[1] Almkvist, G., Krattenthaler, C., & Petersson, J. (2003). Some new formulas for $\pi$. Experimental mathematics, 12(4), 441-456. Link: https://www.arxiv.org/pdf/math.NT/0110238.