Series or product for $\pi$ whose terms are not multiplied by a constant

Question

I'm curious if there exists an infinite sum or infinite product for $\pi$ which is not:

1) a Bailey–Borwein–Plouffe-type formula;
2) of the form $\text{constant}\cdot\sum_{n}a_{n}$ or $\text{constant}\cdot\prod_{n}a_{n}$ (except for the case when the constant is $-1$ or $1$);
3) of the form $f\left(\sum_{n}a_{n}\right)$ or $f\left(\prod_{n}a_{n}\right)$ (except for $f(x)=x$);
4) a sum or product where $a_{n}$ is not elementary.

Some examples of "forbidden" sums and products are $$\pi =\color{red}{4}\displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1}$$ $$\pi =\color{red}{2}\displaystyle\prod_{n=1}^\infty \dfrac{4n^2}{4n^2-1}$$ $$\pi =\color{red}{2}\displaystyle\sum_{n=0}^\infty \dfrac{n!}{(2n+1)!!}$$ $$\pi =\left(\color{red}{\dfrac{1}{16}}\displaystyle\sum_{n=0}^\infty \dbinom{2n}{n}^3\dfrac{42n+5}{2^{12n}}\right)^{\color{red}{-1}}$$ $$\pi =\displaystyle\sum_{n=0}^\infty \dfrac{1}{16^n}\left(\dfrac{4}{8n+1}-\dfrac{2}{8n+4}-\dfrac{1}{8n+5}-\dfrac{1}{8n+5}\right) \quad \color{red}{\text{(BBP-type)}}$$ $$\pi =\displaystyle\sum_{n=1}^\infty \dfrac{3^{n}-1}{4^n}\color{red}{\zeta (n+1)} \quad \color{red}{\text{(non-elementary sequence)}}.$$ For example, numbers such as $e$ or $\ln 2$ satisfy the requirements $1)$, $2)$, $3)$ and $4)$: $$e=\displaystyle\sum_{n=0}^\infty \dfrac{1}{n!}$$ $$\ln 2=\displaystyle\sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n}$$ $$\ln 2=\displaystyle\sum_{n=1}^\infty \dfrac{1}{2^n n}.$$ The formulas for $\pi$ that meet the requirements are, for example (note the "$!$" symbol, since these are nonsensical): $$\pi \overset{!}{=}\displaystyle\sum_{n=1}^\infty \dfrac{2^n}{n^n (1-2n)}$$ $$\pi \overset{!}{=}\displaystyle\prod_{n=1}^\infty \left(1-\dfrac{1}{n^4+1}\right).$$ Hopefully this question is not too vague and you understand what I mean.

Edit: After reading the comments, I must also emphasize that $\sum_{n}\text{constant}\cdot a_{n}$ is "forbidden" as well, since it is the "constant multiplication" in the requirement $2)$. Obviously, cases like $$e=\dfrac{1}{2}\displaystyle\sum_{n=0}^\infty \dfrac{2}{n!}$$ do satisfy the requirements, since the constants cancel out.

Answer 1

$$\pi=3+\sum_1^{\infty}{(-1)^{n+1}4\over(2n+1)^3-(2n+1)}=3+\sum_1^{\infty}{(-1)^{n+1}\over n(n+1)(2n+1)}$$ is Arndt and Haenel 16.10, attributed to Nilakantha, 15th century, with the reference S. Parameswaran, Whish's showroom revisited, The Mathematical Gazette 76 (1992) 28-36. It's also in Borwein and Borwein, Pi and the AGM, page 101.

$$\pi=3+{1\over6+{9\over6+{25\over6+{49\over6+\cdots}}}}$$ is Arndt and Haenel 16.103, attributed to L J Lange, An elegant new continued fraction for $\pi$, American Mathematical Monthly 106 (May 1999) 456-458.

Answer 2

$$\pi=\sum_{-\infty}^{\infty}{\sin n\over n}$$ is formula (16.72) in Arndt and Haenel, $\pi$ Unleashed. It can be derived from formula 508 in Jolley, Summation of Series. Jolley refers to Bromwich, Introduction to the Theory of Infinite Series, page 356.

Arndt and Haenel 16.73 is (using the sinc notation) $$\pi=\sum_{-\infty}^{\infty}{\rm sinc}^2n$$ This can be derived from 520 in Jolley, for which the reference is Whittaker and Watson, Modern Analysis (1920 edition), page 163.

Answer 3

The goal is to find a series for $\pi$ of the form $\sum a_n$, where $a_n$ is elementary, not BBP and not multiplied by a constant (excluding $\pm1$). But this is problematic since the "multiplication" criterion is ill-defined. All $a_n$ can trivially be expressed as a product of $c \cdot \frac{a_n}{c}$, so in that sense they are all multiplied by a constant, hence all $a_n$ are impermissible, so there is no such series. The underlying issue is what remainder is permissible after factoring out the constant. You seem to be implicitly assuming that the remainder must be something like $n^2+1$ or $(-1)^n$, but without other criteria, these are no more natural than the remainder $\frac{1}{2(n!)}$ when $2$ is factored from $\frac{1}{n!}$.

If you just want $a_n$ that can simply be expressed with no constant factor, the series

$$\pi=\sum_{n=0}^\infty\frac{50n-6}{2^n\binom{3n}{n}}$$

suffices (Bailey 2019). As with any other, these terms can be expressed as a multiple of something, since they equal $2\cdot\frac{25n-3}{2^n\binom{3n}{n}}$. Alternatively, you can specify the remainders to be integers and look for a series $\sum\frac{b_n}{c_n}$, such that no constant (except $\pm1$) divides all $b_n$ or $c_n$. For this, you can take Gerry Myerson's series or the series

$$\pi=\sum_{n=1}^\infty\frac{(-1)^{f(n)}}{n}$$

where $f(n)=0$, if $n$ is a prime of the form $4k − 1$; $f(n)=1$, if $n$ is a prime of the form $4k + 1$; $f(n)$ is the sum of $f(i)$ for the factors, $i$ of $n$, if $n$ is composite (Euler 1748). Whether there's an elementary series that fits the criteria, I'm not yet sure.