MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Considering the success of a previous question involving Eulerian numbers, I thought I might throw this question into the mix. It comes from some localization computations in GW theory, but in this form is purely combinatorial.

Eulerian numbers of the second kind are defined by the recursion relation

$$\left\langle\left\langle n\atop m\right\rangle\right\rangle = (m+1)\left\langle\left\langle n-1\atop m\right\rangle\right\rangle+(2n-m-1)\left\langle\left\langle n-1\atop m-1\right\rangle\right\rangle$$

with the initial conditions $\left\langle\left\langle n\atop 0\right\rangle\right\rangle=1$ and $\left\langle\left\langle n\atop m\right\rangle\right\rangle$ = 0 for $m\geq n$. For references, see: and

The following three statements are known:

  1. $\sum_{m=0}^{n} (-1)^mm!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle=0$ for all $n$;

  2. $\sum_{m=0}^{n} (-1)^m(m)m!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle=0$ for odd $n$;

  3. $\sum_{m=0}^{n} (-1)^m(m+1)!(2n-m)!\left\langle\left\langle n\atop m\right\rangle\right\rangle=0$ for even $n$;

(2 and 3 are equivalent).

Question: Show that the expression in the second statement is non-zero for even $n$, i.e. show

$\sum_{m=0}^{n} (-1)^m(m)m!(2n-m-2)!\left\langle\left\langle n\atop m\right\rangle\right\rangle\neq0$ for even $n$.

Certainly we've been checking this on a computer for modest values of $n$.

share|cite|improve this question
Would your question be equivalent to showing that the expression in the third statement is nonzero for odd $n$? – Mike Spivey Nov 12 '10 at 4:57
You may want to ask Manuel Kauers from RISC – Martin Rubey Nov 12 '10 at 9:24
Yes Mike, that question would be equivalent. Call the expression in 2 $A(n)$ and the expression in 3 $B(n)$. Then simply by plugging in the recursion relation, you can show that $A(n)=-B(n-1)$. – Steffen Marcus Nov 12 '10 at 15:56
up vote 10 down vote accepted

Following Mike Spivey's comment above I will consider $B(n)$ in (3). It turns out that your conjecture is true, because for odd $n$ the sum $B(n)$ is a certain weighted $L^2$ norm of the Eulerian polynomial of the second kind of order $\frac{n+1}{2},$ with the sign of $(-1)^{\frac{n-1}{2}}\, . $

The product of the two factorials in $B(n)$ may be expressed in terms of the Eulerian Beta integral $$(m+1)!(2n−m)!=(2n+2)!\int_0^1 t^{m+1 }(1-t)^{2n-m}dt,$$ so that dividing it by $(2n+2)! $ we have $$ \frac {B(n)}{(2n+2)!}= \int_0^1 \sum_{m\ge0}(-1)^m\left\langle\!\!\left\langle n\atop m\right\rangle\!\!\right\rangle t^{m+1 }(1-t)^{2n-m}dt = $$ $$= \int_0^1 t(1-t)^{2n}\sum_{m\ge0} \left\langle\!\! \left\langle n\atop m\right\rangle\!\! \right\rangle \Big(\frac{t}{t-1}\Big)^{ m} dt= \int_0^1 t(1-t)^{2n} E_n \Big(\frac{t}{t-1}\Big) dt. $$ Changing variable with $x:=\frac{t}{t-1}$ this becomes: $$ \int_{-\infty}^0 x(x-1)^{-2n-3} E_n(x) dx, $$

where $E_n$ denotes the Eulerian polynomial of the second kind $$E_n(x):=\sum_{m\ge0} \left\langle\!\!\left\langle n\atop m\right\rangle\!\! \right\rangle x^m,$$ and satisfies the recursive relation (corresponding to the relation for the coefficient that you gave in your question): $$(x-1)^{-2n-2}E_{n+1}(x)=\left( -x(x-1)^{-2n-1}E_n(x) \right)^{\prime}.$$ By the above formula it is now easy to show, integrating by parts repeatedly, that $B(n)=0$ for even $n$ while for odd $n=2p-1$ $$B(2p-1)=(-1) ^ { p + 1 } (4p)! \int_0^{+\infty} E_p (-x)^2 (x+1)^{-4p-1} x dx . $$

(To check this, it is convenient to introduce the sequence of rational functions $U_ n(x):= (x-1)^{-2n}E_n(x)$ that satisfy the recurrence $U_ {n+1}= \big (\frac{x}{1-x} U_ n \big) ^ {\prime} $ with initial condition $U_0=1.$ Hence for all $n+m>0$ we have $\int_{-\infty}^0 U_{n+1}U_{m}\frac{x}{1-x}dx=-\int_{-\infty}^0 U_ {n}U_ {m+1}\frac{x}{1-x}dx\, .$ The integral found above for $B(n) / (2n+2)!\, $ was $-\int_{-\infty}^0 U_{n}U_{1}\frac{x}{1-x}dx$ ).

share|cite|improve this answer
Thanks Pietro! This is very slick and just the sort of thing we've been hoping for... certainly a much more interesting argument than I expected. Thanks also to Mike and Martin for your efforts/feedback. – Steffen Marcus Nov 14 '10 at 23:51
You're welcome! I think you can get further information on the sequence $B(n)$ via a GF. One should start by a GF for the $U_n(x)$, say $u(x,t):=\sum_n U_n(x) t^n$ or $u(x,t):=\sum_n U_n(x) t^n/n!$ that satisfies a PDE coming from the recurrence of the $U_n$ (or another convenient variation). Then $b(t):=\int_{-\infty}^0 u(x,t)x/(1-x) dx$ is essentially a GF for the $B(n)$. – Pietro Majer Nov 15 '10 at 0:24
Interesting... I was initially expecting the solution to be a generating function argument of some type, so your approach was a pleasant surprise. I've been unsuccessful in trying to naively write the generating function for either $A(n)$ and $B(n)$ as a convolution. In fact, they originate from playing with polynomials in the Lambert function. – Steffen Marcus Nov 15 '10 at 2:09

Surprise: your expression is a multiple of a Bernoulli number: $$A(n)=-\frac{(2n)!}{n}B_n $$ for all $n>1$, and the Bernoulli numbers are indeed vanishing exactly for all odd integers greater than 1. This follows computing the generating function for the $A(n)$, on the lines of the preceding answer and comments. Note that your identity can be written in the form $$\sum_{m=0}^n (-1)^n\frac{ \left\langle \!\!\left\langle n\atop m\right\rangle \!\!\right\rangle }{\left( 2n+1\atop m+1\right)}=2B_{n+1}\, .$$

After that, I had a vague memory of a similar relation about the Eulerian numbers of first kind, that it is the fourth identity reported here: $$\sum_{m=0}^n (-1)^n\frac{ \left\langle n\atop m\right\rangle }{\left( n\atop m\right)}=(n+1)B_n\, .$$

Checking the source of the latter may be useful for you, and will possibly give a reference for your identity (I'm quite confident that it should be written somewhere; hopefully some expert may recognize and provide a reference). In case, I will add the details of the (quite standard) computation of the generating function of yours $A(n)$ .

Details. Here is a generating function for yor sequence as a real function. This should allow a nice proof of the Bernoulli number formula, provided one is able to compute the last integral.

For $t\in\mathbb{R}$ and $x\in [0,\infty)$ let $$\psi(t,x):=W(x e^{x+t})=L^{-1}\big(t+L(x)\big)$$ where $W$ is the Lambert function and $L:\mathbb{R_+}\to\mathbb{R}$ is the invertible function $L(x):=x+\log(x)$. So $\psi$ is the general solution of the Cauchy problem for the autonomous ODE (the flow) $$\psi_t=\frac \psi {1+\psi} $$ $$\psi(0,x)=x$$ Thus it also solves the linear first-order PDE $$\psi_t- \frac x {1+x} \psi_x =0$$ and all derivatives $t$ have the form $$\partial_t^n\psi= v_n(\psi)$$ for a recursively defined sequence $v_n(x)$ $$v_0(x)=x$$ $$v_{n+1}(x)=\frac x {1+x}v'_n(x)$$ So the $v_n$ are rational functions with poles at $-1$; $v_n(x)=O(x^{-n})$ as $x\to+\infty$ and in fact $$v_n(x)=-x(1+x)^{-2n+1}E_n(-x)$$ where $E_n$ are the Eulerian polynomial of second kind (these $v_n$ are just a simple modification of to the previously defined sequence $U_n$ ). For all $t\in\mathbb{R}$, all $r>0$ and $n\ge2$, it's easy to see that $$\sup_{|t|\le r}\, \big| v_n\big(\psi(t,x)\big) \big|:=g_{r,n}(x)\in L^1(\mathbb{R}_+),$$ which allows (by the dominated convergence theorem) to differentiate under the sign of integral the function $$h(t):=\int_0^\infty \psi_{tt}(t,x)dx=\int_0^\infty v_2\big(\psi(t,x)\big)dx\, ,$$ so that $$h^{(n)}(t)=\int_0^\infty v_{n+2}\big(\psi(t,x)\big)dx,$$ and in particular we have for $n\ge 1$ $$h^{n-1}(0)=\int_0^\infty v_{n+1}(x)dx=-\frac {A(n)}{(2n)!}$$

The relation $$-\frac {A(n)}{(2n)!} = \frac {B_n} n$$ for all $n\ge 1$ now writes:

$$\int_0^\infty \frac{W(xe^{x+t})}{\big(1+W(xe^{x+t})\big)^3}dx=\frac1 t - \frac 1 {e^t - 1}.$$

share|cite|improve this answer
Very nice! I had a suspicion that Bernoulli numbers were involved somehow, but this is a much simpler expression than I had expected. – Mike Spivey Nov 15 '10 at 18:01
Yes, the world of special numbers is not that wide, after all! ;-) – Pietro Majer Nov 15 '10 at 18:09
This connection is very pleasing to the eye. – Suvrit Nov 16 '10 at 11:14
Thanks again for this. It's quite nice! If it's not too much trouble, would you mind including the details you allude to for the computation of the GF of A(n)? Although I'm sure they are quite standard, as you say, perhaps they are not standard enough for me just yet. I would like to credit you with the identity $A(n)=-\frac{(2n)!}{n}B_n$, citing your answer on mathoverflow of course :-). Obviously, after searching the literature for the last couple of weeks I still haven't been able to find a reference for your identity. – Steffen Marcus Dec 2 '10 at 19:31
Hi, I will add the computation in the next days. I wrote the function $u(x,t)$ defined above, as mentioned there, and obtained a formula in terms of the Lambert function via a linear PDE. Then I integrated over $x \in [-\infty,0]$. Here I made some formal weird simplifications on the integral, and obtained something close to $t/(e^t-1)$, getting the $B_n$. At that point I was completely sure that the last details were standard stuff too, and did not bother to finish the computation, as it semed rather the case of looking for the formula in the literature. – Pietro Majer Dec 3 '10 at 9:09

Your Answer


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.