User:Nathan L. Skirrow/A156744 and square hyperpyramids

Herein, we define $n!=\Gamma (n+1)$ and $H_{n}=\psi (n+1)+\gamma$ for differentiation purposes; with $\sim \!n=-1-n$, the reflection formulae become $n!(\sim \!n)!={\frac {-\pi }{\sin(\pi n)}}$ and $H_{n}-H_{\sim \!n}=-\pi \cot(\pi n)$.

A156744

A156744 is defined such that for any degree-$2n$ polynomial $p$ (with $n\geq 0$), one has

$p'(0)=\sum _{k=-n}^{n}a(n,k)p(k)$

(where $a(n,k)={\frac {A156744(n,k)}{\mathrm {lcm} (1..2n)}}$)

the Lagrange interpolation formula says we have

${\begin{aligned}&p(x)\\=&\sum _{k=-n}^{n}p(k)\prod _{i\in \{-n..n\}\setminus k}{\frac {i-x}{i-k}}\\=&\sum _{k=-n}^{n}p(k){\frac {\frac {(n-x)!}{(k-x)(-1-n-x)!}}{(-1)^{n+k}(n+k)!(n-k)!}}\end{aligned}}$

the summand numerator differentiates to ${\frac {(n-x)!}{(\sim \!n-x)!}}{\frac {{\frac {1}{k-x}}-H_{n-x}+H_{\sim \!n-x}}{k-x}}$, at $x=0$ this becomes ${\frac {n!}{(\sim \!n)!}}{\frac {{\frac {1}{k}}-H_{n}+H_{\sim \!n}}{k}}={\frac {n!^{2}\sin(\pi n)}{-\pi }}{\frac {\pi \cot(\pi n)+{\frac {1}{k}}}{k}}=-n!^{2}{\frac {\cos(\pi n)+{\frac {\sin(\pi n)}{\pi k}}}{k}}$. Since we're concerned with integer indexes in power series, this becomes ${\frac {(-1)^{n+1}n!^{2}}{k}}$, providing the closed form $a(n,k)=[k\neq 0]{\frac {n!^{2}}{k(-1)^{k+1}(n+k)!(n-k)!}}=[k\neq 0]{\frac {(-1)^{k+1}{2n \choose n-k}}{k{2n \choose n}}}$.

even-dimensional square hyperpyramids

we will define the square hyperpyramids slightly differently from Square hyperpyramidal numbers

let $P_{0}(n)=n^{2}$ and $P_{d}(n)=\sum _{k=1}^{n}P_{d-1}(k)={n+d \choose d+1}{\frac {d+2n}{d+2}}$, a figurate number in $d+2$ dimensions. (In that page, $P_{d}(n)=Y_{d+4}^{(d+2)}(n)$) We're concerned with $\sum _{n=1}^{\infty }{\frac {1}{P_{d}(n)}}$.

In particular (note that $(d+1)^{2}{2d+1 \choose d-k}=(d+1)(2d+1){2d \choose d-k}={\frac {(2d+2)!}{2(d-k)!(d+k)!}}$),

${\begin{aligned}&{\frac {1}{P_{2d}(n)}}\\=&{\frac {(-1)^{d}(d+1)^{2}{2d+1 \choose d}}{(d+n)^{2}}}+\sum _{k=-d}^{d}[k\neq 0](-1)^{d+k-1}{\frac {(d+1)^{2}{2d+1 \choose d-k}}{k(d+k+n)}}\\=&{\frac {(-1)^{d}(d+1)^{2}{2d+1 \choose d}}{(d+n)^{2}}}+\sum _{k=1}^{d}(-1)^{d+k-1}{\frac {(d+1)^{2}{2d+1 \choose d-k}}{k}}\left({\frac {1}{d+k+n}}-{\frac {1}{d-k+n}}\right)\\=&\sum _{k=0}^{d}(1+[k>0])(-1)^{d-k}{\frac {(d+1)^{2}{2d+1 \choose d-k}}{(d-k+n)(d+k+n)}}\end{aligned}}$

note that $\lim _{k\rightarrow 0}{\frac {{\frac {1}{d+k+n}}-{\frac {1}{d-k+n}}}{k}}={\frac {-2}{(d+n)^{2}}}$, meaning it can be written without the exception term as $\sum _{k=-d}^{d}(-1)^{d+k-1}{\frac {(d+1)^{2}{2d+1 \choose d-k}}{k(d+k+n)}}$ if you adopt the convention that summands $f(k)$ (when unevaluatable without division by $0$) are to be considered equal to $\lim _{h\rightarrow 0}{\frac {f(k+h)+f(k-h)}{2}}$ (ie. the Cauchy principal value, defined without the integral).

this may be rewritten as ${\frac {(-1)^{d}(d+1)^{2}}{d+n}}{2d+1 \choose d-1}\left({\frac {d+2}{d(d+n)}}+{\frac {{_{3}F_{2}}\left({1,1-d,1-d-n \atop 3+d,2-d-n};1\right)}{1-d-n}}-{\frac {{_{3}F_{2}}\left({1,1-d,1+d+n \atop 3+d,2+d+n};1\right)}{1+d+n}}\right)$

the partial sums are thusly

$\sum _{n=1}^{m}{\frac {1}{P_{d}(n)}}=(-1)^{d}(d+1)^{2}{2d+1 \choose d}\left(H_{d+m}^{(2)}-H_{d}^{(2)}\right)+\sum _{k=-d}^{d}[k\neq 0](-1)^{d+k-1}{\frac {(d+1)^{2}{2d+1 \choose d-k}}{k}}(H_{d+k+m}-H_{d+k})$

since (as $m\rightarrow \infty$) the difference between the $k$th and $-k$th summand approaches $0$, the $H_{d+k+m}$s asymptotically cancel each other out, leaving

$\sum _{n=1}^{\infty }{\frac {1}{P_{d}(n)}}=(-1)^{d}(d+1)^{2}{2d+1 \choose d}\left({\frac {\pi ^{2}}{6}}-H_{d}^{(2)}\right)+\sum _{k=-d}^{d}[k\neq 0](-1)^{d+k}{\frac {(d+1)^{2}{2d+1 \choose d-k}}{k}}H_{d+k}$