# Hyperharmonic number

In mathematics, the n-th hyperharmonic number of order r, denoted by $H_n^{(r)}$, is recursively defined by the relations:

$H_n^{(0)} = \frac{1}{n} ,$

and

$H_n^{(r)} = \sum_{k=1}^n H_k^{(r-1)}\quad(r>0).$

In particular, $H_n=H_n^{(1)}$ is the n-th harmonic number.

The hyperharmonic numbers were discussed by J. H. Conway and R. K. Guy in their 1995 book The Book of Numbers.[1]:258

## Identities involving hyperharmonic numbers

By definition, the hyperharmonic numbers satisfy the recurrence relation

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

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

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

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

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

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

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

## Asymptotics

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

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

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

An immediate consequence is that

$\sum_{n=1}^\infty\frac{H_n^{(r)}}{n^m}<+\infty$

when m>r.

## Generating function and infinite series

The generating function of the hyperharmonic numbers is

$\sum_{n=0}^\infty H_n^{(r)}z^n=-\frac{\ln(1-z)}{(1-z)^r}.$

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

$\sum_{n=0}^\infty H_n^{(r)}\frac{t^n}{n!}=e^t\left(\sum_{n=1}^{r-1}H_n^{(r-n)}\frac{t^n}{n!}+\frac{(r-1)!}{(r!)^2}t^r\, _2 F_2\left(1,1;r+1,r+1;-t\right)\right),$

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.[3]

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

$\sum_{n=1}^\infty\frac{H_n^{(r)}}{n^m}=\sum_{n=1}^\infty H_n^{(r-1)}\zeta(m,n)\quad(r\ge1,m\ge r+1).$

## An open conjecture

It is known, that the harmonic numbers are never integers except the case n=1. The same question can be posed with respect to the hyperharmonic numbers: are there integer hyperharmonic numbers? István Mező proved[4] that if r=2 or r=3, these numbers are never integers except the trivial case when n=1. He conjectured that this is always the case, namely, the hyperharmonic numbers of order r are never integers except when n=1. This conjecture was justified for a class of parameters by R. Amrane and H. Belbachir.[5] Especially, these authors proved that $H_n^{(4)}$ is not integer for all n=2,3,...