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].

reads as

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\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[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,...

External links

Notes

  1. John H., Conway; Richard K., Guy (1995). The book of numbers. Copernicus.
  2. Benjamin, A. T.; Gaebler, D.; Gaebler, R. (2003). "A combinatorial approach to hyperharmonic numbers". Integers (3): 1–9.
  3. 1 2 3 Mező, István; Dil, Ayhan (2010). "Hyperharmonic series involving Hurwitz zeta function". Journal of Number Theory 130: 360–369. doi:10.1016/j.jnt.2009.08.005.
  4. Mező, István (2007). "About the non-integer property of the hyperharmonic numbers". Annales Universitatis Scientarium Budapestinensis de Rolando Eötvös Nominatae, Sectio Mathematica (50): 13–20.
  5. Amrane, R. A.; Belbachir, H. (2010). "Non-integerness of class of hyperharmonic numbers". Annales Mathematicae et Informaticae (37): 7–11.
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