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\bf
\centerline
{
What Is Experimental Mathematics?
}
\rm
\bigskip
\centerline{ {\it
Doron 
ZEILBERGER}\footnote{$^1$}{\eightrm  \raggedright
Department of Mathematics, Rutgers University (New Brunswick),
Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway,
NJ 08854-8019, USA.
%\break
{\eighttt zeilberg  at math dot rutgers dot edu} ,
\hfill \break
{\eighttt http://www.math.rutgers.edu/\~{}zeilberg} .
April 4, 2006.
Transcript of a Colloquium talk given at
Rutgers University, March 10, 2006.
Exclusively published in the Personal Journal of
Ekhad and Zeilberger.
\hfill \break
{\eighttt http://www.math.rutgers.edu/\~{}zeilberg/pj.html} .
Supported in part by the NSF.
}
}

{\bf Abstract:} It is not what you think.

{\bf Why Am I Giving This Talk?}

Thank you all for coming! I really appreciate you showing up.
As for the many colleagues and grad students who did {\it not}
show up, thanks also! Now I won't feel obligated
to come to your talks, or in the case of grad students,
agree to be on your oral committee, or write you a letter of
recommendation.

One of the reasons I agreed to give this
colloquium talk is to find out who are the good guys and who are
the bad guys in this department. Of course, those who are here
(and won't leave early) are the good guys, and their complement
are the bad guys.
I believe that one should 
show up to as many colloquium talks as possible, even if , and in
fact, especially if, they are far removed from one's narrow specialty.
It is like going to Church on Sunday, to Synagogue on Saturday, or
Mosque on Friday. 
My student Eric Rowland will soon take a picture of the audience
and this way I'll know who is here. I promise to always come
to any talk that any of {\it you} will give, or
sponsor. I also promise {\it not} to come to any talk given
or sponsored by faculty or grad students
who are not here right now, unless it is really really interesting.

So that was the first reason for giving this talk. The
second reason is to illustrate {\bf by example}, how to give
a perfect colloquium talk, accessible to everyone.
The third reason is to illustrate the {\bf Zeilberger methodology}
of teaching Freshman Calculus.
The main ingredients of my method are
the {\bf handout} and the {\bf ``quiz''}.
In each and every one of my classes,
I give out a handout, explaining in excruciating detail how
to do the homework problems from the covered section.
Then, in the last five minutes, everyone has to attempt a ``quiz''
that does not count towards the grade, except that it is used to
check attendance (attendance is strictly mandatory in my classes).
I ask my students to write their name and E-mail address, and for those
who didn't get it perfectly right, I write them an E-mail explaining
what they did wrong. I know that you are not freshmen, 
but still,
just to illustrate my method, I will give out a ``quiz''
at the end of my talk and will write you E-mail on Monday
in case you didn't get it right. 

Last, but not least, the fourth reason  for giving this talk is to
tell you about experimental mathematics the way I and
my students practice it. It is not the conventional way!

\vfill\eject

{\bf The Experimental Math Group at Rutgers}

Speaking of my students, together with Hill Assistant Professor
Drew Sills, and myself, they comprise the Experimental
Mathematics Group here at Rutgers. 
We have a weekly seminar, that meets at Hill 705 every Thursday
between 5:00pm and 5:50pm, that has (usually) exciting talks.
You are all welcome to come as often as you wish.

Let me describe
very briefly what they are up to. For more details, see their
websites.

{\bf Drew Sills} is one of the greatest experts in the world
in RogersRamanujanology and developed a great package,
{\tt RRtools}, for exploring (and proving!) fascinating results
in that theory. In collaboration with me, we taught the computer
how to automatically  discover and {\it prove} far-reaching
extensions of the celebrated Dyson Constant Term Identity
(first proved by physics Nobelist Ken Wilson).
{\bf Moa Apagodu} does beautiful extensions and
applications of WZ theory. {\bf Lara Pudwell}
and my recently graduated student {\bf Vince Vatter}  teach computers
how to enumerate permutations much more efficiently (and rigorously!)
then any human can do. {\bf Eric Rowland} does, among other things,
Cellular Automata, in the style of Stephen Wolfram. 
{\bf Thotsaporn Thanatipanonda} (a.k.a. ``Aek'') teaches
computers how to play {\it Chess} but, unlike Deep Blue and
Deep Junior, not on an $8 \times 8$ board (this is not math!)
but on an $m \times n$ board. He was able to find better
winning strategies, even for $m=n=8$, then 
those suggested by former world champion
Jose Capablanca in his classic book. 
For now, of course, he does simple endgames,
but I am sure that he will be able to do increasingly more complicated
positions. I should also mention {\bf Arvind Ayyer}, whose work is
not exactly experimental mathematics, but it is very interesting
nevertheless, applying combinatorial identities to string theory.

{\bf Finally, What is Experimental Math?}

Let me try and explain to you, by example, what is
experimental math. It is really an {\it attitude} and
{\it way of thinking}, or rather {\it not thinking}.
Mathematicians traditionally love to solve problems by
{\bf thinking}. Myself, I hate to think. I love
to {\bf meta-think}, try to do things, whenever possible, by
{\bf brute force}, and of course, let the computer do the
hard work.

I just came back from a conference in Florida, and one of the
other plenary speakers, the famous Michigan number theorist
{\bf Hugh Montgomery}, who is also an avid problem-solver
(and hence a closet combinatorialist) told me about
a cute problem that he proposed for the Putnam exam-
on whose committee he is currently serving-
and that was rejected by the other members of the committee
for being ``too hard''. He wrote it down on a napkin\footnote
{$^2$}
{\eightrm
It was during the  reception on March 6, 2006,
of the 37th International Southeastern
Conference on Combinatorics, Graph Theory
and Computing held at Florida Atlantic University.
On the same napkin, Hugh Montgomery scribbled
the following challenging problem.
``Let $I$ and $J$ be intervals on the real line, and let
$q$ and $r$ be integers, $q \geq 1$.
If $|I| \cdot |J| <q$ then
$\{ n \in I : rn \in J (mod \,\, q) \}$ is
a segment of an arithmetical progression.''
This interesting problem was
mailed to Montgomery,
in the mid seventies, from a cabin in Montana,
by the Michiganer alumnus T.J. Kaczynski.
}
[the napkin is shown to the audience], and here it is:

{\bf Hugh Montgomery's Rejected Putnam Proposal}

Show that there exists a polynomial $P \in R[x,y]$, $P \neq 0$, such
that $P(F_n,F_{n+1})=0$ for every $n$, where $F_n$ are the
Fibonacci numbers defined by $F_0=0,F_1=1$ and
$F_n=F_{n-1}+F_{n-2}$ for $n \geq 2$.

{\bf The Experimental Math Approach}

Go to Maple and type

\tt
with(combinat):
P:=(d,x,y)->add(add(a[i,j]*x**i*y**j,i=0..d-j),j=0..d);

V:=d->$\{$seq(seq(a[i,j],i=0..d-j),j=0..d)$\}$;

E:=d->$\{${seq(P(d,fibonacci(n),fibonacci(n+1)),n=1..nops(V(d))+5) $\}$:

Q:=(d,x,y)->subs(solve(E(d),V(d)),P(d,x,y));

\rm

Now type {\tt Q(1,x,y);} and get $0$, which is
correct, but does not count, since Hugh Montgomery insisted on
a {\it non-zero} polynomial. Then we try, in turn,
{\tt Q(2,x,y);} and {\tt Q(3,x,y);} and still get $0$.
But type {\tt Q(4,x,y);},  {\it et voil\`a}, Maple returns
the answer
$$
(x^2+xy-y^2-1)(x^2+xy-y^2+1) \quad,
$$
implying the {\it Fibonacci identity} relating two consecutive
Fibonacci numbers
$$
(F_n^2+F_nF_{n+1}-F_{n+1}^2-1)(F_n^2+F_nF_{n+1}-F_{n+1}^2+1)=0 \quad.
$$

Now most of you would say, `that's very nice, but it is only
a conjecture, we still need humans to prove it for {\it all} n'. Nonsense!
Once discovered, any polynomial identity (and many other types!)
involving Fibonacci numbers, like the one above, is routinely and
automatically provable by computer. Why? (hint: Binet's formula).

Now to do this, it took me one minute to type the Maple code
(since I am a slow typist), and took the computer less than a second
to discover (and prove) the above polynomial identity. In fact,
in this case (and in many other cases), there is no need for
a proof, using, say, Binet's formula alluded to in the paragraph above.
By general linear algebra nonsense it is easy to show that if
the identity is true for sufficiently many special cases, it is
true for every $n$, and since {\it E(d)} was defined with a few
equations to spare, everything is watertight.

I am not sure how Hugh Montgomery solved this problem, but
I am sure that it took him more than  a minute.

For the sake of the Maple illiterate, let me explain what
the code means. $P(d,x,y)$ defines a generic polynomial
of $x$ and $y$ of degree $d$, with {\it undetermined coefficients}.
$$
P(d,x,y):=\sum_{j=0}^{d} \sum_{i=0}^{d-j} a_{i,j} x^iy^j \quad,
$$
while $V(d)$ is the set of coefficients:
$$
V(d):=\{a_{i,j} \quad \vert \quad i \geq 0, j \geq 0, i+j \leq d \} \quad.
$$
$E(d)$ is the set of equations
$$
E(d):=\{P(F_n,F_{n+1})=0 \quad \vert \quad 1 \leq n \leq |V(d)|+5 \} \quad .
$$
Finally, {\tt solve(E(d),V(d))} solves the system of
$|V(d)|+5$ equations  and $|V(d)|$ unknowns, and
$Q(d,x,y)$ substitutes the general solution into
$P(d,x,y)$.

{\bf Experimental Mathematics through the ages}

Until relatively recently, about $2500$ years ago,
{\it all} of mathematics was experimental. 
The sages of ancient Egypt, Sumer,
Babylon, India and China had a very advanced empirical science.
Then came Euclid and his cronies (who had an axe to grind 
with Zeno and other skeptics) and invented
the axiomatic method, and turned mathematics into
a {\it deductive} science, at least officially.
Of course, all the great mathematicians (and also
the not so greats) {\it implicitly} did experiments
(until sixty years ago, with pencil and paper
[or finger and sand, but watch out for Roman soldiers]),
but most of them covered their traces and didn't tell anyone.

{\bf Leonhard Euler: Experimental Mathematician}

One of the few giants who did not make any bones about
empirical discoveries was the great Euler. 
Possibly because of his friend Goldbach, he was interested
in a characterization of prime numbers. He may have also
had in mind the problem of proving that there are infinitely
many perfect numbers and that there are no odd perfect numbers.
Now a good way to characterize both {\it prime} and
{\it perfect} numbers is via the {\it sum of divisors} function
$\sigma(n)$
$$
\sigma(n)=\sum_{d|n} d \quad .
$$
For example $\sigma(6)=1+2+3+6=12$. Now $p$ is a prime
iff $\sigma(p)=p+1$ and $n$ is perfect iff $\sigma(n)=2n$.
So if you know everything about $\sigma(n)$ you would be able
to do Goldbach, Twins, and Perfect.

The first item on the agenda is to {\it compute} the first
thousand, or so, values of $\sigma(n)$. Since factoring is
hard, it would be nice if one could only use addition
and subtraction, and be able to use previously
computed values of $\sigma$. So Euler tried to find
some kind of recurrence. Here is how he reasoned. First
he used what Herb Wilf calls {\it generatingfunctionology}.
Obviously, if $d(n)$ is the {\it number of divisors} of $n$,
then
$$
\sum_{n=0}^{\infty} d(n) x^n=
\sum_{i=1}^{\infty} {{x^i} \over {1-x^i}} \quad,
$$
since the $x^i/(1-x^i)$ contributes one term to each of the
multiples of $i$. Now for the {\it sum} of divisors, one has
similarly
$$
\sum_{n=0}^{\infty} \sigma(n) x^n=
\sum_{i=1}^{\infty} {{ix^i} \over {1-x^i}} \quad .
$$
But the right side is essentially the logarithmic derivative of
$(1-x^i)$, so we have
$$
\sum_{n=0}^{\infty} \sigma(n) x^n=
-x\sum_{i=1}^{\infty} [\log (1-x^i)]'=
-x \left ( \log \left [ \prod_{i=1}^{\infty} (1-x^i)] \right ] \right )' =
-x {{\Pi'} \over {\Pi}} \quad,
$$
where 
$$
\Pi=\prod_{i=1}^{\infty} (1-x^i) \quad .
$$
Now, just with pencil and paper, Euler expanded the first 
forty or whatever terms of
this infinite product to get
$$
\Pi=1-x-x^2+x^5+x^7-x^{12}-x^{15}+x^{22}+x^{26}-x^{35}-x^{40}+ etc.,
$$
[Euler used `etc.' instead of $\dots$],
and he immediately recognized the exponents as {\bf pentagonal numbers}, and
made the

{\bf Conjecture}:
$$
\prod_{i=1}^{\infty} (1-x^i) =1+\sum_{j=1}^{\infty} (-1)^j [ x^{j(3j-1)/2}+x^{j(3j+1)/2} ] \quad,
$$
but couldn't find a {\bf demonstration}. He said that he had a {\it preuve},
by which he meant an empirical proof, but lacked a {\it demonstration}, meaning
a formal proof. Nevertheless, being a true experimental mathematician, it
did not stop him from drawing important consequences, and he postponed the
task of finding a complete proof to when he'll get around to it.

Now, he  multiplied both sides of
$$
\sum_{n=0}^{\infty} \sigma(n) x^n=
-x {{\Pi'} \over {\Pi}} \quad,
$$
by $\Pi$,  getting
$$
\Pi \sum_{n=0}^{\infty} \sigma(n) x^n=-x \Pi'  \quad,
$$
and assuming his conjecture for now, and comparing coefficients of $x^n$ on both sides
he derived a recurrence for $\sigma(n)$:
$$
\sigma(n)=\sum_{j=1}^{[\sqrt{2n/3}]} (-1)^{j-1} [ \sigma(n-j(3j-1)/2)+\sigma(n-j(3j+1)/2) ] \quad,
$$
if $n$ is not a pentagonal number, and if $n$ is a pentagonal number 
$n=k(3k \pm 1)/2$, then
$$
\sigma(n)=\sum_{j=1}^{[\sqrt{2n/3}]} (-1)^{j-1} 
[ \sigma(n-j(3j-1)/2)+\sigma(n-j(3j+1)/2) ] +(-1)^{k+1} n \quad ,
$$
that enabled Euler to compile a table of $\sigma(n)$ for $n \leq N$ in time  $O(N^{3/2})$.

Only twenty five years later,  when he was more then seventy years old, 
did he find
a formal proof. Of course, he did lots of other things in between, 
but we should learn from him not to get hung up on one step.


{\bf Carl Friedrich Gauss: Experimental Mathematician}

Also the great Guass was an excellent experimental mathematician,
although you wouldn't know it from his {\it published} work.
But his fascinating diary contains lots of experimentations.
For example, consulting tables of prime numbers, he guessed
the {\bf Prime Number Theorem} that $\pi(n)$, the number of
primes less than $n$ is asymptotically $n/\ln n$:
$$
\pi(n) \sim {{n} \over {\ln n}} \quad .
$$

{\bf Srinivasa Ramanujan: The Greatest Experimental Mathematician}

Srinivasa Ramanujan had an amazing intuition on what is interesting
and correct and he conjectured lots and lots of results that took many
mathematician-hours to prove formally, and some of his discoveries
are still being proved today by Bruce Berndt, George Andrews, and their
disciples. Some of his results he proved himself, after G.H. Hardy
tried to teach him about `proof', but I am sure that he was only
trying to be nice to Hardy, and he had no doubt that his 
discoveries were correct, even without a formal proof.

One of his most famous results, that
$$
p(5k+4) \equiv 0 \, (mod \,\,\, 5) \quad,
$$
where $p(n)$ is the {\it number of integer partitions} of $n$
was discovered when he looked at the table of this sequence
computed by Percy MacMahon using the recurrence
$$
p(n)=\sum_{j=1}^{[\sqrt{2n/3}]} (-1)^{j-1} 
[ p(n-j(3j-1)/2)+p(n-j(3j+1)/2) ] \quad ,
$$
that is very similar, but slightly simpler then Euler's recurrence
above for $\sigma(n)$. It also follows from Euler's Pentagonal Number
Theorem (even simpler, there is no need to take the logarithmic derivative),
as was first  observed by Euler.

\vfill\eject
{\bf The Computer Age}

Many parts of modern math, for example {\it fractals} and {\it chaos}
wouldn't be possible without experimentation done on the computer.
This is obvious. However, I know of quite a few important
theorems in very pure math that were discovered by computer
experimentation, but the ungrateful human author never even
acknowledged this fact!  

{\bf Mitchell Feigenbaum}

Playing with a programmable calculator that his parents bought
him as a graduation present, a young postdoc 
at Los Alamos, in the early seventies,
was iterating the mapping
$$
x \rightarrow \lambda x(1-x) \quad , \quad (0 < x <1)
$$
for various values of $\lambda$, and noticed that for small values
of $\lambda$ it tends to just one limiting value
(regardless of the starting number), then suddenly, in the long
run, it alternates between two limiting values, then $4$, then
$8$, etc, until chaos rules completely.
Not only that, calling the cutoffs values of $\lambda$,
$\lambda_1$, $\lambda_2$, $\lambda_3$, etc., Feigenbaum
noticed that
$$
\lim_{n \rightarrow \infty} { { \lambda_{n+1}-\lambda_{n}}
                             \over
                             { \lambda_{n}-\lambda_{n-1}}}
\,\, = \, \, 4.669\dots \quad ,
$$

the {\it now} famous Feigenbaum constant. The rest is history.


{\bf Bailey-Borwein-Plouffe}

There are lots of formulas for $\pi$, some converge extremely slowly,
like Leibnitz's
$$
\pi= 4 \sum_{n=0}^{\infty} {{(-1)^n} \over {2n+1}} \quad,
$$
and some very fast. But the following beautiful formula,
$$
\pi=\sum_{n=0}^{\infty} 
\left ( 
{{4} \over {8n+1}}-
{{2} \over {8n+4}}-
{{1} \over {8n+5}}-
{{1} \over {8n+6}} \right )
{{1} \over {16^n}} \quad ,
$$
due to David Bailey, Peter Borwein, and Simon Plouffe is
special, since it enables one to compute the zillionth and first
binary digit of $\pi$ without having to first compute
the previous zillion digits. Once known, its proof is an easy calculus
exercise, but {\it discovering} it, using the methodology
of {\it experimental mathematics}, was a major tour-de-force.

If you want to know more about `mainstream' experimental mathematics
read the wonderful two-volume set on this subject
written by David Bailey and Jon Borwein (v.1)
and Bailey, Borwein  and  Roland Girgensohn (v.2).

\vfill\eject

{\bf Zeilberger-style Experimental Mathematics}

Traditionally there was a {\bf dichotomy} between the
{\bf context of discovery}, that nowadays is mostly done
by computers, and the {\bf context of verification} that
is still mostly carried out by humans. In my style of experimental
math, the computer does {\it everything},
the {\it guessing} and the (rigorous!) {\it proving}, if possible
completely seamlessly without any human intervention.
Feel free to browse my website for many examples.

{\bf Handout Time}

As I promised, I will now hand out a {\bf handout}, that
illustrates the methodology in terms of a very simple
example, that of the classical {\bf gambler's ruin}.

\bigskip
\hrule
\medskip
\centerline
{
\bf Dr. Z's March 10 Colloquium Handout
}

{\bf Problem Type}: A function is given by some rule.
First conjecture, and then rigorously prove, an explicit
formula for it.


%%General Problem


{\bf Example Problem}: Let $f_L(n)$ be the expected life
of a gambler in a fair casino, who currently has $n$ dollars
and loses or wins a dollar with probability $1/2$, and has
to quit when he is ruined, or when he has $L$ dollars.
Find an explicit formula for $f_L(n)$ as a function of
$n$ and $L$.

%%Specific Problem


\bigskip
\hrule
\bigskip
%\halign{#\hfill  & \quad #\hfill \cr
\halign{           
\vtop{\hsize=15pc\noindent #\strut} \qquad
&
\vtop{\hsize=15pc\noindent #\strut} \cr
{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} 
%%%General Step 1
Using human (or computer) ingenuity, set up an
algorithm for computing the function numerically
for ``small'' values of the arguments.

%%%End General Step 1
&
%%%Specific Step 1
{\bf 1.} 
For a fixed $L$, the $n+1$ quantities $f_L(n)$ ($n=0 \dots L$)
are completely characterized by the system of $n+1$ 
linear equations with $n+1$  unknowns
$$
f_L(n)={{1} \over {2}} f_L(n-1)+
{{1} \over {2}} f_L(n+1)+1 \quad , \quad ( 1 \leq n \leq L-1)
$$
$$
f_0(0)=0 \quad, \quad f_L(L)=L \quad .
$$



%%%End Specific Step 1
\cr
\bigskip & \bigskip \cr
{\bf 2.} 
%%% General Step 2
Using a linear solver, crank out enough data.
%%% End General Step 2
&
{\bf 2.} 
%%%Specific Step 2
$$
f_3(0)=0, f_3(1)=2, f_3(2)=2, f_3(3)=0 \quad ,
$$
$$
f_4(0)=0, f_4(1)=3, f_4(2)=4, f_4(3)=3, f_4(4)=0 \quad ,
$$
$$
f_5(0)=0, f_5(1)=4, f_5(2)=6, f_5(3)=6, f_5(4)=4, f_5(5)=0 \quad .
$$ 
%%%End Specific Step 2

\cr \bigskip & \bigskip \cr

{\bf 3.} 
%%% General Step 3
Conjecture an {\it ansatz} and use 
{\it undetermined coefficients} to find it
(if possible).

%%% End General Step 3
&
{\bf 3.} 
%%% Specific Step 3
Setting $f_L(n)=an^2+bn+c$ and solving for $a,b,c$
(for each specific $L$) we get
$$
f_3(n)=n(3-n) \quad f_4(n)=n(4-n) \quad f_5(n)=n(5-n) 
$$
$$
\quad f_6(n)=n(6-n)  \quad, \quad f_7(n)=n(7-n)  \quad.
$$
%%% End Specific Step 3

\cr
\bigskip & \bigskip \cr
{\bf 4.} 
%%% General Step 4
Using the same methodology conjecture an expression in $L$
for $f_L(n)$, then prove it {\it rigorously} by 
using the {\it defining algorithm},
but now {\bf symbolically}.
%%% End General Step 4
&
{\bf 4.} 
%%%Specific Step 4
$$
f_L(n)=n(L-n) \quad .
$$
Now let's prove it! Let's denote \hfill\break
$g_L(n):=n(L-n)$. We have
$$
g_L(n)-{{1} \over {2}} g_L(n-1)- {{1} \over {2}} g_L(n+1)-1 =
$$
$$
L(n-L)-{{1} \over {2}} (n-1)(L-(n-1))- {{1} \over {2}} (n+1)(L-(n+1))-1 =0 
\quad .
$$
Also $g_L(0)=0 \cdot (L-0)=0$ and $g_L(L)=L(L-L)=0$.
So we have a {\bf rigorous} proof that $f_L(n)=g_L(n)$ for
{\it every} $n$ and $L$.
%%%End Specific Step 4

\cr \bigskip & \bigskip \cr

\cr}
\hrule
\vfill\eject

\medskip
{\bf Quiz Time}

As I promised you, now you have a chance to test your understanding.
Here is the ``quiz''. I am sure that most of you will get it
right, but in case you don't, I'll be writing you an
E-mail on Monday, explaining your errors.
\bigskip
\hrule
\bigskip
\centerline
{
\bf ``QUIZ'' for March 10, 2006, Dr. Z.'s Colloquium talk
}
\bigskip
{\bf NAME:} (print!) $\underline
{\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

\bigskip
{\bf E-MAIL ADDRESS:} (print!) 
$\underline
{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$
\bigskip

{\bf 1.} Conjecture and prove an explicit expression for
$$
a(n):=\sum_{i=1}^{n} i \quad
$$
using the methodology of {\bf experimental mathematics}
(no credit for other methods!) .

{\bf Hint:} $a(n)$ is characterized by $a(n)-a(n-1)=n$ and
$a(0)=0$.

\bigskip
\hrule
\bigskip
{\centerline \bf APPENDIX}

{\bf Who handed in the quiz?}

Arvind Ayyer, Dick Bumby, Greg Cherlin, ``Deep Purple'',
Sue Geller [visiting from Texas A\&M],
Zheng-Chao Han, Jim Lepowsky, Laurie Moore, 
Roger Nussbaum, Amit Priyadarshi, Lara Pudwell,
Eric Rowland, Drew Sills, 
Thotsaporn Thanatipanonda (``Aek''),
Roderich Tumulka [visiting from Tuebingen],
Jerrold Tunnel, Liming Wang, Chuck Weibel, Robert Wilson.


You all did a great job! Thanks.

{\bf Answer to the Quiz}

By direct calculation $ a(0)=0, a(1)=1, a(2)=3, a(3)=6, a(4)=10$.
Guess a quadratic $a(n)= \alpha n^2 + \beta n + \gamma$
and get the system of equations
$$
\{ \gamma=0, \alpha+\beta+\gamma=1, 
4\alpha+2\beta+\gamma=3, 
9\alpha+3\beta+\gamma=6,
16\alpha+4\beta+\gamma=10 \} \quad,
$$
in the set of variables $\{ \alpha, \beta, \gamma \}$.
Solving this set of equations yields
$\alpha= 1/2$, $\beta=1/2$ and $\gamma=0$, that leads us to
{\bf conjecture} that $a(n)=(1/2)n^2+(1/2)n$.
Let $b(n):=(1/2)n^2+(1/2)n$.
Now $b(0)=0$ and
$$
b(n)-b(n-1)=
(1/2)n^2+(1/2)n- [ (1/2)(n-1)^2+(1/2)(n-1)]= n \quad .
$$
Since both $x(n)=a(n)$ and $x(n)=b(n)$ satisfy
$x(0)=0$ and $x(n)-x(n-1)=n$ it follows that
$a(n) \equiv b(n)$ . \halmos 

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