Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In my other thread I discussed a matrix-decomposition; for one matrix (U) I found now a description of its entries, which may best be denoted as "recursive harmonic numbers". However, googling with this term leads to hits on a different thing. What I have is the following.

Define a recursive type of harmonic-numbers of two arguments:

$ \qquad \small h_{0,c} =1 \text{ for all } c\gt 0 $
$ \qquad \small h_{r,1} =1 \text{ for all } r\gt 0 $
$ \qquad \small h_{r,c} = h_{r,c-1} + h_{r-1,c}/c \text{ for all } c\gt r $

Then we have
$\qquad \small h_{0,*} = (1,1,1,1,1,\ldots) $
$\qquad \small h_{1,*} = (1, 1+{1 \over 2}, 1+{1 \over 2}+{1 \over 3}, 1+{1 \over 2}+{1 \over 3}+{1 \over 4}, \ldots) $ $\qquad \small h_{2,*} = (1, 1+ {1 \over 2}\left( 1+{1 \over 2}\right), 1+ {1 \over 2}\left(1+{1 \over 2}\right) +{1 \over 3}\left(1+{1 \over 2}+{1 \over 3},\right), \ldots) $
where the recursion becomes visible by
$\qquad \small h_{2,*} = (1, 1+ {1 \over 2} h_{1,2}, 1+ {1 \over 2} h_{1,2} + {1 \over 3} h_{1,3} \ldots) $

Is someone aware of such a generalization/extension? I have a vague idea that I've seen the terms of $\small h_{2,*} $ in some online available article not too long ago (arXiv?) but can't remember exactly. Also with "generalized harmonic numbers" mostly is meant either the summation to fractional bounds of the index and/or higher exponents in the denominator of the harmonic numbers, so this is different...


I've found the arXiv-article. However, the recursive harmonic numbers are only mentioned as intermediate terms (pg 6, eq (6)) for a generalization of the Stirling numbers first kind (which is in the context of my own discussion not surprising). If there are some more references dealing more specific with these, this were very good...

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

Your recursion can be expressed as $$h_{r,c} = \sum_{i=1}^c \frac{h_{r-1,i}}{i},$$ which, unrolled, becomes the nested sum $$h_{r,c} = \sum_{i_r = 1}^c \frac{1}{i_r} \left(\sum_{i_{r-1} = 1}^{i_r} \frac{1}{i_{r-1}} \left(\cdots \sum_{i_1=1}^{i_2} \frac{1}{i_1} \right)\right) = \sum_{i_r = 1}^c \sum_{i_{r-1} = 1}^{i_r} \cdots \sum_{i_1=1}^{i_2} \frac{1}{i_1 i_2 \cdots i_r}.$$ If we reverse the order of summation, we get $$h_{r,c} = \sum_{i_1 = 1}^c \sum_{i_2 = i_1}^c \cdots \sum_{i_r=i_{r-1}}^c \frac{1}{i_1 i_2 \cdots i_r},$$ which is almost identical to an expression I asked about on MO a while back. A slight modification of Richard Stanley's answer there shows that $$h_{r,c} = Z_c(H_c, H^{(2)}_c, \ldots, H^{(c)}_c),$$ where $Z_c(x_1, x_2, \ldots, x_c)$ is the $c$th cycle index polynomial of the symmetric group. In other words, your $h_{r,c}$ has the same expression in terms of the generalized harmonic numbers $H^{(r)}_n$ as my $S(n,k)$ except that $h_{r,c}$ has all of the terms positive. For instance, we have

$$\begin{align} h_{2,c}&= \frac{1}{2}\left(H_c^2 + H^{(2)}_c \right), \\ h_{3,c} &= \frac{1}{6}\left(H_c^3 + 3H_c H^{(2)}_n + 2 H_c^{(3)}\right),\\ h_{4,c} &= \frac{1}{24}\left(H_c^4 + 6 H^2_c H_c^{(2)} + 3 (H_c^{(2)})^2 + 8 H_c H_c^{(3)} + 6 H_c^{(4)}\right). \end{align}$$

See also the references in my question.

The $h_{r,c}$ values can also be expressed in terms of the generalized Stirling numbers in the Loeb paper you cite. See, for example, the recursion (3) in Loeb's paper, which is the same as that for $h_{r,c}$. It's just the initial conditions that are different. Specifically, $h_{r,c} = (-1)^r c! s(-c,r)$, in which case Proposition 2.3 in Loeb's paper says that $$h_{r,c} = -\sum_{m=1}^c \binom{c}{m} (-1)^m m^{-r}.$$

share|improve this answer
    
Mike, again thanks for your answer. Just for the record: I've collected my findings so far in a small article, where unfortunately the (infinite) set of identities involving the Stirling numbers first kind could not be confirmed in completeness - I only found solutions for individual entries of the W-matrix (see my other thread). For convenience - here is the link to my compilation: go.helms-net.de/math/divers/inverseNullmatrix.pdf –  Gottfried Helms Dec 15 '11 at 17:22
    
@Gottfried: I'm glad it was helpful. And thanks for the link. –  Mike Spivey Dec 15 '11 at 19:11
add comment

(This is more of an extended comment.)

I'll only note that your $h_{r,c}$ are easily generated as the first column of the product of a diagonal matrix and a lower triangular matrix. More precisely, the matrix that generates your $h_{r,c}$ can be expressed as $\mathbf D\mathbf L^{r+1}$, where $\mathbf D=\mathrm{diag}(1,2,\dots,n)$ and

$$\ell_{j,k}=\begin{cases}\frac1{j}&\text{if }j\geq k\\0&\text{otherwise}\end{cases}$$

$\mathbf L$, which is apparently called the "lower triangular mean matrix", looks a bit like this:

$$\mathbf L=\begin{pmatrix}1\\\frac12&\frac12\\\frac13&\frac13&\frac13\\\vdots&\vdots&\vdots&\ddots\\\frac1{n}&\cdots&\cdots&\cdots&\frac1{n}\end{pmatrix}$$

$h_{2,c}$ is expressible in terms of generalized harmonic numbers: $\dfrac{(H_c)^2+H_c^{(2)}}{2}$, where $H_c^{(k)}=\sum\limits_{j=1}^c\frac1{j^k}$. I haven't been able to find a simpler form for $h_{3,c}$ and others.

share|improve this answer
    
Very nice, thanks! Interestingly, the matrix L has a simple eigen-decomposition, which allows then fractional/even complex powers. Immediately we've found some fractional generalizations for the "recursive harmonic numbers"... The eigenmatrices are the lower triangular Pascal-matrix and its inverse, and the eigenvalues are the obvious ones. We can even compare that generalizations with that of D.Loeb and others... cool –  Gottfried Helms Dec 12 '11 at 16:46
    
Hmm, the term "... mean matrix" in OEIS is not much explained. I'd prefer the reminder, that it is the matrix for the Hölder-summation : consider a summable sequence to be summed in a column-vector A, then P=D*L*A contains the partial sums, and with k'th powers of L leftmultiplied $\small H_k = L^k \cdot P $ gives the Höldersum of k'th order (of a possibly divergent sum) –  Gottfried Helms Dec 12 '11 at 16:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.