# Generalized series for $\pi$ - What is the polynomial?

The Madhava-Leibniz series for $$\pi$$ is $$4\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$$

When we regroup odd and even terms into individual new terms, $$b_n=a_{2n}+a_{2n+1}$$, we have the other well known series for $$\pi$$

$$8\sum_{n=0}^\infty\frac{1}{(4n+1)(4n+3)}$$

But then continuing this and iteratively regrouping the terms of each series gives us a sequence of series

$$2^{k+1}\sum_{n=0}^\infty\frac{p_k(n)}{(2^kn+1)(2^kn+3)(2^kn+5)\ldots}$$

for $$k=1,2,\ldots$$, where $$p_k(n)$$ is (for $$k>1$$) a polynomial in $$n$$ and where the product in the denominator has terms $$(2^kn+i)$$, for odd $$1\leq i<2^k$$.

From the relation $$b_n=a_{2n}+a_{2n+1}$$, we get a recursive definition of $$p_k$$ that can be simplified into factorials

$$p_{k+1}\left(n\right)=\frac{1}{2^{\left(2^{\left(k-1\right)}+1\right)}}\left(\frac{\left(2h\right)!i!}{h!\left(2i\right)!}p_k(2n)+\frac{\left(2i\right)!w!}{i!\left(2w\right)!}p_k(2n+1)\right)$$

where $$w=2^kn,\ i=2^{\left(k-1\right)}+2^kn,\ h=2^k\left(n+1\right)$$. The first few $$p_k$$ are $$\begin{array}{|c|c|} \hline k&p_k\\ \hline 1&(-1)^n\\ \hline 2&1\\ \hline 3&64n^2+64n+19\\ \hline 4&16777216n^6+\ldots+191115\\ \hline \end{array}$$

So, my question is this: is there an explicit, general closed form for $$p_k(n)$$? If so, what is it and has it been studied? Also, is there a name for the series regrouping $$b_n=a_{2n}+a_{2n+1}$$?

The reason I was initially interested in this rearrangement is because this SE question calls for a series with terms that are elementary functions but lack a constant factor. Hypothetically, a generalised series could be extended to $$k=-1$$, which would omit the constant $$2^{k+1}$$.

• But with $k=-1$ you would have $2^{k-1}=1/2$ terms at the denominator! Also, "each" of them would be of the form $(n/2+i)$, thus with an implicit constant factor. – mbjoe Sep 1 '19 at 19:35
• $1/4$ terms sorry. – mbjoe Sep 1 '19 at 19:51
• @mbjoe Indeed, that may mean it's impossible to generalise the series for $k<1$. But we can't know unless we try, and the end result may be interesting :). Sometimes functions can be generalized in ways that would have initially been counterintuitive (e.g. $\sum n^{-s}$ being generalised to the Riemann zeta function on $\mathbb{C}$). For now I'm just focusing on the smaller question of 'what is $p(n)$?'. – Jam Sep 1 '19 at 19:56