Hyperharmonic number

From Wikipedia, the free encyclopedia
Jump to: navigation, search

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

Notes[edit]

  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. ^ a b c Mező, István; Dil, Ayhan (2010). "Hyperharmonic series involving Hurwitz zeta function". Journal of Number Theory (130): 360–369. 
  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.