Hyperharmonic number

By definition, the hyperharmonic numbers satisfy the recurrence relation

${\displaystyle H_{n}^{(r)}=H_{n-1}^{(r)}+H_{n}^{(r-1)}.}$

In place of the recurrences, there is a more effective formula to calculate these numbers:

${\displaystyle H_{n}^{(r)}={\binom {n+r-1}{r-1}}(H_{n+r-1}-H_{r-1}).}$

The hyperharmonic numbers have a strong relation to combinatorics of permutations. The generalization of the identity

${\displaystyle H_{n}={\frac {1}{n!}}\left[{n+1 \atop 2}\right].}$

reads as

${\displaystyle H_{n}^{(r)}={\frac {1}{n!}}\left[{n+r \atop r+1}\right]_{r},}$

where ${\displaystyle \left[{n \atop r}\right]_{r}}$ is an r-Stirling number of the first kind.^[2]

The above expression with binomial coefficients easily gives that for all fixed order r>=2 we have.^[3]

${\displaystyle H_{n}^{(r)}\sim {\frac {1}{(r-1)!}}\left(n^{r-1}\ln(n)\right),}$

that is, the quotient of the left and right hand side tends to 1 as n tends to infinity.

An immediate consequence is that

${\displaystyle \sum _{n=1}^{\infty }{\frac {H_{n}^{(r)}}{n^{m}}}<+\infty }$

when m>r.

Summarize Timeline Fact Check

The generating function of the hyperharmonic numbers is

${\displaystyle \sum _{n=0}^{\infty }H_{n}^{(r)}z^{n}=-{\frac {\ln(1-z)}{(1-z)^{r}}}.}$

The exponential generating function is much more harder to deduce. One has that for all r=1,2,...

${\displaystyle \sum _{n=0}^{\infty }H_{n}^{(r)}{\frac {t^{n}}{n!}}=e^{t}\left(\sum _{n=1}^{r-1}H_{n}^{(r-n)}{\frac {t^{n}}{n!}}+{\frac {(r-1)!}{(r!)^{2}}}t^{r}\,_{2}F_{2}\left(1,1;r+1,r+1;-t\right)\right),}$

where ₂F₂ is a hypergeometric function. The r=1 case for the harmonic numbers is a classical result, the general one was proved in 2009 by I. Mező and A. Dil.^[4]

The next relation connects the hyperharmonic numbers to the Hurwitz zeta function:^[3]

${\displaystyle \sum _{n=1}^{\infty }{\frac {H_{n}^{(r)}}{n^{m}}}=\sum _{n=1}^{\infty }H_{n}^{(r-1)}\zeta (m,n)\quad (r\geq 1,m\geq r+1).}$