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Bellard's "exotic" pi formula 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 $m=7$. However, it seems there are an infinite number of such formulas for other orders $m$.

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. (Edited) 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:

$$740025\pi+20379280 = \sum_{n=1}^\infty \dfrac{3P_1(n)}{{\displaystyle \tbinom{7n}{2n}2^{n-1}}}\tag{2a}$$

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

where,

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

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

  1. The eqn $(2a)$ is Bellard's, but I found there are also $P(n)$ that are $6$th, $7$th deg, and so on, with different $a,b$.
  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)

$$7997795704284513820875\pi+186851093786889785568000= \sum_{n=1}^\infty \dfrac{P(n)}{{\displaystyle \tbinom{11n}{2n}2^{n-1}}}\tag3$$

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 $a,b$.
  2. The prime factors of $a=7997795704284513820875$ are, $$3, 5, 7, 13, 17, 19, 23, 37, 41, 43, 59, 61, 79.$$

II. Comment

  1. I found these using Mathematica's integer relations sub-routine. It could not find similar identities for order $m=4v+1=5,9,13$ with proportionately-sized coefficients, indicating that these may be only for $m = 4v+3 = 3,7,11,\dots$

III. Questions

  1. Can one find identities for all $m = 4v+3 = 3,7,11,\dots$?
  2. Is there really an infinite number of identities per order $m$?
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  • $\begingroup$ What exactly is the reason behind the existence of these expressions ? Are the polynomials in question unique for each n $($save multiplication by a constant, of course$)$ ? $\endgroup$
    – Lucian
    Jun 19, 2015 at 7:14
  • $\begingroup$ @Lucian: You mean for each order $m$? Apparently the polynomials $P(n)$ are unique. The one for $m=3$ seems easiest to tackle since there are only four terms to track, so there might be a closed-form. I surprised Bellard didn't check $m=7$ if there were other $P(n)$ that would do, nor why he didn't check $m=11$. (Must be speed constraints since he found it back in 1997.) $\endgroup$ Jun 19, 2015 at 7:21
  • $\begingroup$ Is Bellard's work published anywhere? $\endgroup$ Jun 19, 2015 at 7:28
  • $\begingroup$ @GerryMyerson: Only in Mathworld and Bellard's website. I don't think it has been rigorously proven yet. $\endgroup$ Jun 19, 2015 at 7:31
  • $\begingroup$ @TitoPiezasIII: Ah, yes... $1997$... those were the days ! :-$)$ $\endgroup$
    – Lucian
    Jun 19, 2015 at 7:34

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