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

Define recursively polynomials $f_n(a,b)$ by $$ f_0(a,b)=1,\ \ f_n(0,b)=0\ \mathrm{for}\ n>0 $$ $$ \frac{\partial}{\partial a}f_n(a,b) = f_{n-1}(b-a,1-a). $$ For instance, $$ f_1(a,b) = a,\ \ f_2(a,b) = \frac 12(2ab-a^2) $$ $$ f_3(a,b) = \frac 16(a^3-3a^2-3ab^2+6ab). $$ Is there a ``nice'' solution to this recurrence, e.g., a formula for the generating function $\sum_{n\geq 0}f_n(a,b)x^n/n!$? What I am really interested in is $f_n(1,1)$. For the motivation, see the solution to Exercise~4.56(d) (pg. 645) of Enumerative Combinatorics, vol.1, 2nd ed.

share|cite|improve this question
Hi Richard, I added the page number of the solution; I hope you don't mind. – Suvrit Feb 7 '12 at 22:20
Here are the values of $f_n(1,1)$ for $0\le n\le 15$: $1,1,1,1,2,5,14,47,182,786,3774,19974,115236,720038,4846512,34950929$ (of course they don't mach anything in the OEIS). – Pietro Majer Feb 8 '12 at 0:25
You mean $n!f_n(1,1)$. – Richard Stanley Feb 8 '12 at 1:07
(yes sorry, I meant the numerators) – Pietro Majer Feb 8 '12 at 9:01
It is in the OEIS: A096402, submitted by a certain Stanley ;). – Brendan McKay Feb 8 '12 at 10:43

There seems to be a PDE for $g(a,b,x)=\sum_{n\ge0}f_n(a,b)x^n$, which can be thought of as a boundary value problem in the triangle $0\lt a\lt b\lt1$. $$g_{aab}+g_{abb}+x^3g=0$$ ($x$ is a parameter and subscripts are derivatives) with boundary values $g(0,b,x)=1$, $g_a(a,a,x) = x$, and $g_{ab}(a,1,x) = x^2$. This comes from iterating the $f_n$ recurrence, after Pietro's remarks that $(a,b)\to(b-a,1-a)$ has period 3 suggested looking at third derivatives. Does that determine $g$ uniquely, nicely? I don't know yet. [Edit: I wrongly wrote $g$ at first using $\frac{x^n}{n!}$.]

share|cite|improve this answer

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.