# Hyperharmonic number

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

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

and

${\displaystyle H_{n}^{(r)}=\sum _{k=1}^{n}H_{k}^{(r-1)}\quad (r>0).}$[citation needed]

In particular, ${\displaystyle H_{n}=H_{n}^{(1)}}$ 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

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

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

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

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

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

## Asymptotics

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.

## Generating function and infinite series

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]

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

## 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[5] 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.[6] Especially, these authors proved that ${\displaystyle H_{n}^{(4)}}$ is not integer for all r<26 and n=2,3,... Extension to high orders was made by Göral and Sertbaş.[7] These authors have also shown that ${\displaystyle H_{n}^{(r)}}$ is never integer when n is even or a prime power, or r is odd.