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 $\pi$


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


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 $2^{k+1}$.

  • $\begingroup$ 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. $\endgroup$ – mbjoe Sep 1 '19 at 19:35
  • $\begingroup$ $1/4$ terms sorry. $\endgroup$ – mbjoe Sep 1 '19 at 19:51
  • $\begingroup$ @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 $p(n)$?'. $\endgroup$ – Jam Sep 1 '19 at 19:56

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