%oddarmin.tex: %On the Intriguing Problem of Counting (n+1,n+2)-core partitions into Odd Parts %begin macros \def\L{{\cal L}} \baselineskip=14pt \parskip=10pt \def\halmos{\hbox{\vrule height0.15cm width0.01cm\vbox{\hrule height 0.01cm width0.2cm \vskip0.15cm \hrule height 0.01cm width0.2cm}\vrule height0.15cm width 0.01cm}} \font\eightrm=cmr8 \font\sixrm=cmr6 \font\eighttt=cmtt8 \magnification=\magstephalf \def\P{{\cal P}} \def\Q{{\cal Q}} \def1円{{\overline{1}}} \def2円{{\overline{2}}} \parindent=0pt \overfullrule=0in \def\Tilde{\char126\relax} \def\frac#1#2{{#1 \over #2}} %\headline={\rm \ifodd\pageno \RightHead \else \LeftHead \fi} %\def\RightHead{\centerline{ %Title %}} %\def\LeftHead{ \centerline{Doron Zeilberger}} %end macros \centerline {\bf On the Intriguing Problem of Counting $(n+1,n+2)$-core partitions into Odd Parts } \rm \centerline {\it Anthony ZALESKI and Doron ZEILBERGER} {\bf Abstract}: Tewodros Amdeberhan and Armin Straub initiated the study of enumerating subfamilies of the set of $(s,t)$-core partitions. While the enumeration of $(n+1,n+2)$-core partitions into {\it distinct} parts is relatively easy (in fact it equals the Fibonacci number $F_{n+2}$), the enumeration of $(n+1,n+2)$-core partitions into {\it odd} parts remains elusive. Straub computed the first eleven terms of that sequence, and asked for a ``formula," or at least a fast way, to compute many terms. While we are unable to find a ``fast" algorithm, we did manage to find a ``faster" algorithm, which enabled us to compute 23 terms of this intriguing sequence. We strongly believe that this sequence has an algebraic generating function, since a ``sister sequence" (see the article), is OEIS sequence A047749 that does have an algebraic generating function. One of us (DZ) is pledging a donation of 100 dollars to the OEIS, in honor of the first person to generate sufficiently many terms to conjecture (and prove non-rigorously) an algebraic equation for the generating function of this sequence, and another 100 dollars for a rigorous proof of that conjecture. Finally, we also develop algorithms that find explicit generating functions for other, more tractable, families of $(n+1,n+2)$-core partitions. {\bf Added Jan. 24, 2018}: Paul Johnson observed ([J]) and proved that the "sister-sequence" counts $(n+1,n+2)$-core partitions into {\bf even} parts, and even more impressively, related the two sequences, that easily implies a fast way to compute the enumerating sequence that we were interested in, from which one can easily derive a (rather complicated!) algebraic equation satisfied by the generating function. We would have needed 53ドル$ terms (rather than 23ドル$) to have guessed it. {\bf Added Feb. 28, 2018}: Paul Johnson has just posted his beautiful article {\it Simultaneous cores with restrictions and a question of Zaleski and Zeilberger}, {\tt https://arxiv.org/abs/1802.09621} \quad , that does much more than we asked. A donation to the OEIS, of \200,ドル in honor of Paul Johnson, has been made. {\bf Supporting Maple Packages and Output} All the results in this article were obtained by the use of the Maple packages $\bullet$ {\tt http://www.math.rutgers.edu/\~{}zeilberg/tokhniot/OddArmin.txt} \quad , $\bullet$ {\tt http://www.math.rutgers.edu/\~{}zeilberg/tokhniot/core.txt} \quad , $\bullet$ {\tt http://www.math.rutgers.edu/\~{}zeilberg/tokhniot/stCorePlus.txt} \quad , whose output files, along with links to diagrams, are available from the {\it front} of this article {\tt http://www.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimhtml/oddarmin.html} \quad . {\bf $(s,t)$-Core Partitions} Recall that a {\it partition} is a non-increasing sequence of positive integers $\lambda=(\lambda_1, \dots, \lambda_k)$ with $k \geq 0,ドル called its {\it number of parts}; $n:=\lambda_1 + \dots + \lambda_k$ is called its {\it size}, and we say that $\lambda$ is a {\it partition of $n$}. Also recall that the {\it Ferrers diagram} (or equivalently, using empty squares rather than dots, {\it Young diagram}) of a partition $\lambda$ is obtained by placing, in a {\it left-justified} way, $\lambda_i$ dots at the $i$-th row. For example, the Ferrers diagram of the partition $(5,4,2,1,1)$ is $$ \matrix { * & * & * &* & * \cr * & * & * &* & \cr * & * & & & \cr * & & & & \cr * & & & & \cr } \quad . $$ Recall also that the {\it hook length} of a dot $(i,j)$ in the Ferrers diagram, 1ドル\leq j \leq \lambda_i,ドル is the number of dots to its right (in the same row) plus the number of dots below it (in the same column) plus one (for itself), in other words $\lambda_i -i+\lambda'_j-j+1,ドル where $\lambda'$ is the {\it conjugate partition}, obtained by reversing the roles of rows and columns. (For example if $\lambda=(5,4,2,1,1)$ as above, then $\lambda'=(5,3,2,2,1)$.) Here is a table of hook-lengths of the above partition, $(5,4,2,1,1)$: $$ \matrix { 9 & 6 & 4 & 3 & 1 \cr 7 & 4 & 1 &1 & \cr 4 & 1 & & & \cr 2 & & & & \cr 1 & & & & \cr } \quad . $$ It follows that its set of hook-lengths is $\{1,2,3,4,6,7,9\}$. A partition is called an $s$-core if none of its hook-lengths is $s$. For example, the above partition, $(5,4,2,1,1),ドル is a 5ドル$-core, and an $i$-core for all $i \geq 10$. A partition is a {\it simultaneous} $(s,t)$-core partition if it avoids hook-lengths of both $s$ and $t$. For example, the above partition, $(5,4,2,1,1),ドル is a $(5,8)$-core partition (and a $(5,10)$-core partition, and a $(100,103)$-core partition etc.) For a lucid and engaging account, see [AHJ]. As mentioned in [AHJ], Jaclyn Anderson ([An]) very elegantly proved the following. {\bf Theorem ([An])} If $s$ and $t$ are relatively prime positive integers, then there are {\bf exactly} $$ \frac{(s+t-1)!}{s!t!} \quad $$ $(s,t)$-core partitions. For example, here are the $(3+5-1)!/(3!5!)=7$ $(3,5)$-core partitions: $$ \{ empty , 1, 2, 11, 31, 211, 4211 \} \quad . $$ {\bf The Order Ideal $P_{n+1,n+2}$} It turns out that it is most convenient to work with the order ideal $$ P_{s,t} := {\bf N} \backslash (s{\bf N} + t{\bf N} ) \quad, $$ where ${\bf N}$ is the set of non-negative integers. Anderson showed that $(s,t)$-core partitions are in one-to-one correspondence with {\it order ideals} of $P_{s,t}$ ([An]). Our poset of interest, $P_{n+1,n+2},ドル can be identified with a triangular region in the 2D rectangular lattice, let's call it $A_n,ドル $$ A_n:=\{ (i,j) \in {\bf N}^2 ,円 \vert ,円 0 \leq i \leq n-1 ,円 , ,円 0 \leq j \leq n-1-i \} \quad, $$ consisting of $(n+1)n/2$ lattice points. We {\bf label} the lattice point with $$ L(i,j) := (n+1)n-1 - (n+2)i - (n+1)j \quad, $$ in other words, we identify the lattice point $(i,j)$ with the member $(n+1)n-1 - (n+2)i - (n+1)j$ of $P_{n+1,n+2}$. To see the lattice $P_{9,10}$ see {\tt http://sites.math.rutgers.edu/\~{}zeilberg/tokhniot/PictOddArmin/O0.html} \quad , and to see the lattice $P_{10,11}$ see {\tt http://sites.math.rutgers.edu/\~{}zeilberg/tokhniot/PictOddArmin/O1.html} \quad . Note that the point $(n-1,0)$ is labeled 1ドル,ドル and when we read the labels along diagonals, from the bottom-right to the top-left, the labels increase by 1ドル,ドル but as we move from the end of one diagonal to the next one there are ``discontinuities'' of sizes 3,4,ドル \dots,n+1$ respectively. A subset $S$ of $A_n$ is an {\it order ideal} if it satisfies the following condition: $\bullet$ If $(i,j) \in S$ then $(i',j') \in S$ for {\it all} $(i',j') \in S$ such that $i' \geq i $ and $j' \geq j$. In other words, if a lattice point belongs to $S,ドル then so do all the lattice points of $A_n$ that are {\it both} to its (weak) right {\it and} are (weakly) above it. Conversely, and equivalently, $\bullet$ If $(i,j) \not \in S,ドル then the set $\{(i',j') \in A_n ,円 | ,円 i' \leq i \quad and \quad j' \leq j \}$ is disjoint from $S$. In other words, if a lattice point is unoccupied, then all the lattice points (weakly) below and (weakly) to its left are also unoccupied. With this interpretation, it is very easy to prove the special case of Anderson's theorem that the number of $(n+1,n+2)$-core partitions is Catalan's number $(2n+2)!/((n+1)!(n+2)!)$. Given such an order ideal, let's call it $S,ドル let $i$ (0ドル \leq i \leq n$) be the smallest positive integer with the property that $(n-1-i,i)$ is {\bf not} a member of $S$: in other words, the smallest integer $i$ such that $(n-1,0),(n-2,1), \dots, (n-i,i-1)$ are members of $S$ while $(n-1-i,i)$ is {\bf not} a member of $S$. Then all the points to the left of and below $(n-1-i,i)$ are definitely unoccupied, and the order ideal has two parts, $\bullet$ Those (strictly) below and (strictly) to the right of $(n-1-i,i)$ \quad , $\bullet$ Those (strictly) above and (strictly) to the left of $(n-1-i,i)$ \quad . The former component is isomorphic to an order ideal of $A_{i-1}$ with its outer diagonal fully occupied; i.e., by definition of $i,ドル $\{(n-1,0), (n-2,1), \dots, (n-i,i-1)\}$ are occupied, and removing these ``mandatory" members, the remaining set is isomorphic to an order ideal of $A_{i-2}$. Let's call it $S_1$. The top part is isomorphic to an order ideal of $A_{n-i},ドル let's call it $S_2$. This introduces a {\bf canonical decomposition} $$ S \rightarrow (i, S_1, S_2) \quad, \quad 0 \leq i \leq n \quad , \quad S_1 \in A_{i-2} \quad , \quad S_2 \in A_{n-i} \quad \eqno{(Canonical Decomposition)} $$ that is obviously one-to-one. Let $a_n$ be the number of order ideals of $A_n$; then it follows that it satisfies the recurrence $$ a_n ,円 = ,円 \sum_{i=0}^{n} a_{i-1} a_{n-i-1} \quad, $$ with the initial condition $a_{-1}=1$. As is well-known (and easy to see) this implies that indeed $a_n=(2n+2)!/((n+1)!(n+2)!)$. Note that for the above argument the labels are irrelevant. {\bf Counting subfamilies} {\bf Distinct Partitions} What about counting $(n+1,n+2)$-core partitions into {\bf distinct} parts? It was conjectured by Amdeberhan[Am], and first proved (as a special case of a much more general result) by Straub[S1] that this number is $F_{n+2},ドル where $(F_n)$ is the Fibonacci sequence.. Using order ideals, this is immediate. By the transformation $$ (a_1, a_2, \dots , a_k) \rightarrow (a_1-(k-1), \dots, a_k) \quad, $$ from the sorted list of labels to an $(n+1,n+2)$-core partition, the condition of being distinct translates to the fact that the corresponding order ideal can't have any adjacent points, when ``read" along diagonals, from right-to-left and from bottom-to-top. This precludes any member of an inner diagonal (since their existence would imply at least two adjacent points in the outermost diagonal), and the members of $S$ that do belong to the outer-diagonal can't be adjacent. Let $d_n$ be their number. If $(n-1,0)$ is not a member of $S$ then $S$ can be viewed as an order ideal (with the above condition) of $A_{n-1},ドル accounting for $d_{n-1}$ such creatures. On the other hand, if $(n-1,0) \in S,ドル then $(n-2,1) \not \in S ,ドル and removing both of these yields an order ideal of $A_{n-2}$ (with the above conditions), hence $d_n$ satisfies the recurrence $$ d_n = d_{n-1} + d_{n-2} \quad , $$ subject to the initial conditions $d_{0}=1,ドル $d_{1}=2$. For multi-cores see the elegant paper [AmL]. {\bf Intermezzo: The Joy and Agony of Enumerative Combinatorics} It is both fascinating and frustrating that in enumeration problems, `tweaking' a problem ever so slightly turns it from almost trivial (and often, utterly trivial) to very difficult (and often, intractable). For example, it is utterly trivial that the number of $n$-step walks in the 2D rectangular lattice is 4ドル^n,ドル but just add the adjective ``self-avoiding"---in other words, the number of such walks that never visit the same vertex twice---and the enumeration problem becomes (most probably) intractable and, at any rate, wide open. Another example is counting permutations that avoid a pattern. The number of permutations, $\pi,ドル of length $n$ that avoid the pattern 12ドル$ (i.e. you can't have 1ドル \leq i_1

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