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{\bf  
An Irrationality measure of arctan(1/p)
}
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{\it Doron ZEILBERGER and Wadim ZUDILIN}
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{\bf Abstract.} We establish an irrationality measur for $\arctan(\frac{1}{p})$ for any odd prime $p$.

{\bf The Maple package.} This article is accompanied by a shoft Maple package

$\bullet$ {\tt ArcTan.txt}  \quad ;

It is available, along with an input and output file, from the front of this article

{\tt http://www.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimhtml/arctan.html} \quad .

Let $p$ be an odd prime. Consider the following integral
$$
E_n(p) \, = \, \int_0^1 \, \frac{x^n\,(1-x)^n}{(1+\frac{x^2}{p^2})^{n+1}}\, dx \quad .
$$

{\bf Fact 1}: There exists a constant $C_1$ such that
$$
|E_n(p)| \,  \leq \, \frac{C}{a_1(p)^n} \quad,
$$
where
$$
a_1(p) \, = \, 2\,{\frac {1}{ \left( -p+\sqrt {1+{p}^{2}} \right) p}} \quad.
$$

{\bf Proof}: Differentiating $f(x)= \frac{x(1-x)}{1+\frac{x^2}{p^2}}$ with respect to $x$ and setting it equal to $0$
gets that the maximum is at  $x=\left( -p+\sqrt {1+{p}^{2}} \right) p$. Plugging it into $f(x)$ gives
$1/a_1(p)$.


{\bf Fact 2}: $E_n(p)$ satisfies the linear recurrence equation 
$$
(4n+8)E_{n+2}(p) - 2p^2 (2n+3) E_{n+1}(p) -(n+1)p^2 E_{n}(p)=0 \quad .
$$


{\bf Proof}: Follows from the Almkvist-Zeilberger algorithm [AZ].

{\bf Fact 3}: There exist rational numbers $A_n(p)$ and $B_n(p)$ such that
$$
E_n(f) \, = \, -A_n \,+ \,B_n \, arctan(\frac{1}{p}) \quad .
$$

{\bf Proof}: Either by integration by parts, or using Fact 2, once it is verified for $n=0$ and $n=1$.


{\bf Fact 4}: The sequence of rational numbers $A_n(p)$, $B_n(p)$ satisfy the same recurrence as of $E_n(p)$, namely

$$
(4n+8)A_{n+2}(p) + 2p^2 (2n+3) A_{n+1}(p) -(n+1)p^2 A_{n}(p)=0 \quad , \quad
$$
$$
(4n+8)B_{n+2}(p) + 2p^2 (2n+3) B_{n+1}(p) -(n+1)p^2 B_{n}(p)=0 \quad.
$$

{\bf Proof}: Use fact 3, comparing the coefficients of $1$ and $\arctan(\frac{1}{p})$.

{\bf Fact 5}: There exist constants $C_2$, $C_3$ such that
$$
A_n(p) \leq  C_2 a_2(p)^n \quad , \quad
B_n(p) \leq  C_3 a_2(p)^n \quad ,
$$
where
$$
a_2(p) \, = \, \frac{1}{2}\, \left( p+\sqrt {1+{p}^{2}} \right) p
$$

{\bf Proof}: From Fact 4 and the Poincar\'e Lemma.


{\bf Fact 6}: There exists a constant $C_4$ such that
$$
|arctan(\frac{1}{p}) \, - \, \frac{A_n(p)}{B_n(p)}| \leq \frac{C_4}{(a_1(p) \, a_2(p))^n} \quad .
$$

{\bf Proof}: Divide Fact 3 by $B_n$ and use Facts 1 and 5.


{\bf Surprising Fact 7}: Let $d(n):=lcm(1...n)$. Then
$$
\bar{A}_n(p) = A_n(p) \, d(n) \, 2^{[3n/2]}/p^n \quad , \quad 
\bar{B}_n(p) = B_n(p) \, d(n) \, 2^{[3n/2]}/p^n \quad 
$$
are integers.

{\bf Proof}: Checked empirically for many $p$ and many $n$. Formal proof coming up.

{\bf Fact 8}:
$$
\bar{B}_n(p) \, = \, O( (2^{3/2}\,a_2(p)/p)^n d(n)) \quad ,
$$

{\bf Proof}: From Fact 5 and the definition of $\bar{B}_n(p)$.

From Fact 6 we have

{\bf Fact 9}
$$
|arctan(\frac{1}{p}) \, - \, \frac{\bar{A}_n}{\bar{B}_n}| \leq \frac{C_4}{(a_1(p) \, a_2(p))^n} =\frac{C_5}{(\bar{B}_n(p))^{1+\delta}} \quad,
\quad ,
$$
where (since $d(n)=O(e^n)$),
$$
1+ \delta = \frac{\log(a_1(p))+ \log (a_2(p))}{\frac{3}{2} \log 2 + \log(a_2(p))-\log(p)+1} \quad.
$$

As usual (see [vdP]), the irrationality measure is $1+ 1/\delta$, hence, we have established

{\bf Theorem}: If $p$ is an odd prime, then $arctan(\frac{1}{p})$ has irrationality measure
$$
-2\, \left( \ln  \left( 2\,{\frac {p+\sqrt {1+{p}^{2}}}{p}} \right) +\ln  \left( 1/2\, \left( p+\sqrt {1+{p}^{2}} \right) p \right)  \right)  \left( -2\,
\ln  \left( 2\,{\frac {p+\sqrt {1+{p}^{2}}}{p}} \right) +3\,\ln  \left( 2 \right) +2-2\,\ln  \left( p \right)  \right) ^{-1} \quad.
$$

Here are a few sample values

$$
p=3: 7.707357546 \quad, \quad
p=5: 4.788327432 \quad, \quad
p=7: 4.075567842 \quad, \quad
p=11:  3.542000040 \quad, \quad
$$
$$
p=1009: 2.429989300 \quad , \quad
p =1000000007:   2.134188776  \quad , \quad
p=100000000000000000039: 2.059322676 \quad .
$$

{\bf References}

[AZ] Gert Almkvist and Doron Zeilberger, {\it The Method of differentiating Under The integral sign},
J. Symbolic Computation {\bf 10} (1990), 571-591. Available from \hfill\break
{\tt http://sites.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimhtml/duis.html } \quad .

[PWZ] Marko Petkovsek, Herbert S. Wilf, and Doron Zeilberger, {\it ``A=B''}, A.K. Peters, 1996. Freely available from 
{\tt https://www.math.upenn.edu/\~{}wilf/AeqB.html} \quad .

[vdP] Alfred van der Poorten, {\it A Proof that Euler missed}, Math. Intell. {\bf 1}(1979), 195-203.

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Wadim Zudilin, Institute for Mathematics, Astrophysics and Particle Physics
Radboud Universiteit, PO Box 9010
6500 GL Nijmegen, The Netherlands
Email:  {\tt  wzudilin [at] gmail [dot] com} \quad .
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Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen
Rd., Piscataway, NJ 08854-8019, USA. \hfill\break
Email: {\tt DoronZeil [at] gmail  [dot] com}   \quad .
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Written: Oct. 8, 2019.
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