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While trying to get an expression for the number of deals from a generalised bridge deck with nobody being void in any suit I encountered the following subproblem.

From a generalised bridge deck with $r$ ranks instead of just $13$ deal four hands of $r$ cards each. How many deals $A(r)$ are there where

  • South is void in $\diamondsuit$ and $\heartsuit$
  • West is void in $\clubsuit$ and $\heartsuit$
  • North is void in $\clubsuit$ and $\diamondsuit$
  • East is unrestricted, though all players may be void in other suits?

$A(r)$ is the coefficient of $(wxyz)^r$ in $((w+z)(x+z)(y+z)(w+x+y+z))^r$ where $wxyz$ correspond to $\clubsuit\diamondsuit\heartsuit\spadesuit$ respectively. I solved this as follows:

  • South picks $a$ spades, West picks $b$ spades and North picks $c$ spades in one of $\binom r{a,b,c,r-a-b-c}$ ways
  • East gives $r-a$ clubs, $r-b$ diamonds and $r-c$ hearts to South, West, North respectively in one of $\binom ra\binom rb\binom rc$ ways; the rest of the deal is forced

Thus $$A(r)=\sum_{a,b,c}\binom ra\binom rb\binom rc\binom r{a,b,c,r-a-b-c}$$ which after applying Vandermonde's identity to $c$ followed by shuffling binomials becomes the double sum over all $a,b$ $$A(r)=\sum_{a,b}\binom ra^2\binom ab^2\binom{r+b}a$$ But when I computed the terms and compared to the OEIS I found a match with A290575, with a much simpler formula $$B(r)=\sum_{k=0}^r\binom rk^2\binom{2k}r^2$$ and I cannot prove that these two binomial sums are equal.

How can I show that $A=B$, preferably (but definitely not necessarily) with a combinatorial proof? I know of Zeilberger's algorithm but don't have access to Maple.

Parcly Taxel
  • 105,904

1 Answers1

1

The following holonomic proof follows Peter Paule and Carsten Schneider's 2024 report Creative Telescoping for Hypergeometric Double Sums, and was generated using their software.

Denote the inner sum $$f(r,a)=\sum_{b=0}^a\binom ab^2\binom{r+b}a=\sum_{b=0}^as(r,a,b)$$ and compute two recurrences for it:

In[1]:= << RISC`fastZeil`
<< Downloads/Sigma.m

In[5]:= sd = Binomial[a, b]^2 Binomial[r + b, a];
prerod = Zb[sd, {b, 0, a}, a];
rod = ReleaseHold[prerod[[1]]] /. SUM -> f


Out[5]= -(1 + a)^2 f[a] - (3 + 2 a) (1 + 2 r) f[1 + a] + (2 + a)^2 f[2 + a] == 0


In[7]:= {r0, r1, r2} = FunctionExpand[{sd, (sd /. r -> r + 1), (sd /. a -> a + 1)}/sd];
prehook = Gosper[sd, {b, 0, a}, Parameterized -> {r0, r1, r2}];
hook = ReleaseHold[prehook[[1, 1, 1]]] == 
   0 /. {DisplayForm[SubscriptBox["F", "0"]][b] -> f[a], 
   DisplayForm[SubscriptBox["F", "1"]][b] -> f[r + 1, a], 
   DisplayForm[SubscriptBox["F", "2"]][b] -> f[a + 1]}


Out[7]= (-1 - a^2 - 2 r + 2 a r - 2 r^2) f[a] - (1 + a)^2 f[1 + a] + 
  2 (1 + r)^2 f[1 + r, a] == 0

$$(a+1)^2f(r,a)+(2a+3)(2r+1)f(r,a+1)-(a+2)^2f(r,a+2)=0$$ $$2(r+1)^2f(r+1,a)-(a^2-2ar+2r^2+2r+1)f(r,a)-(a+1)^2f(r,a+1)=0$$ The first recurrence has certificate (obtained using Prove[]) $$g(r,a,b)=\frac{(a+1)b^2(b-a+r)(2a^2-3ab-3ar+5a+b^2+2br-4b-6r+2)}{(b-a-2)^2(b-a-1)^2}s(r,a,b)$$ in the sense that the following telescoping equation holds: $$(a+1)^2s(r,a,b)+(2a+3)(2r+1)s(r,a+1,b)-(a+2)^2s(r,a+2,b)=g(r,a,b+1)-g(r,a,b)$$ Dividing both sides by $s(r,a,b)$ results in an easily verifiable rational function identity. In exactly the same manner the second recurrence is certified by $$g(r,a,b)=-\frac{b^2(a^2-ab-3ar-a+2br+b-3r-2)}{(a-b+1)^2}s(r,a,b)$$ Using these two recurrences we now compute one for $A(r)$:

In[9]:= CreativeTelescoping[
 SigmaSum[Binomial[r, a]^2 f[a], {a, 0, r}], r, {{rod, f[a]}}, hook]

Out[9]= {{(1 + r)^3, -(1/4) (3 + 2 r) (7 + 9 r + 3 r^2), 
  1/16 (2 + 
     r)^3, ((80 + 8 a - 206 a^2 + 206 a^3 - 104 a^4 - 10 a^5 + 
        10 a^6 + 456 r + 160 a r - 1095 a^2 r + 843 a^3 r - 
        214 a^4 r - 25 a^5 r + 7 a^6 r + 1020 r^2 + 658 a r^2 - 
        2182 a^2 r^2 + 1172 a^3 r^2 - 160 a^4 r^2 - 12 a^5 r^2 + 
        1158 r^3 + 1114 a r^3 - 2054 a^2 r^3 + 696 a^3 r^3 - 
        44 a^4 r^3 + 708 r^4 + 916 a r^4 - 924 a^2 r^4 + 
        152 a^3 r^4 + 222 r^5 + 364 a r^5 - 160 a^2 r^5 + 28 r^6 + 
        56 a r^6) f[a] Binomial[r, a]^2)/(32 (-2 + a - r)^2 (-1 + a - r)^2) + ((1 + 
        a)^2 (-80 + 152 a - 98 a^2 - 10 a^3 + 10 a^4 - 296 r + 
        448 a r - 213 a^2 r - 5 a^3 r + 7 a^4 r - 428 r^2 + 
        486 a r^2 - 156 a^2 r^2 + 2 a^3 r^2 - 302 r^3 + 230 a r^3 - 
        38 a^2 r^3 - 104 r^4 + 40 a r^4 - 14 r^5) f[1 + a] Binomial[r, a]^2)/(32 (-2 + a - r)^2 (-1 + a - r)^2)}}

Call the gnarly fourth element of the output $h(r,a)$: $$ \frac{\binom ra^2}{32(r-a+2)^2(r-a+1)^2} \left(\begin{bmatrix} 80 & 8 & -206 & 206 & -104 & -10 & 10 \\ 456 & 160 & -1095 & 843 & -214 & -25 & 7 \\ 1020 & 658 & -2182 & 1172 & -160 & -12 \\ 1158 & 1114 & -2054 & 696 & -44 \\ 708 & 916 & -924 & 152 \\ 222 & 364 & -160 \\ 28 & 56 \end{bmatrix}f(r,a) + (1+a)^2\begin{bmatrix} -80 & 152 & -98 & -10 & 10 \\ -296 & 448 & -213 & -5 & 7 \\ -428 & 486 & -156 & 2 \\ -302 & 230 & -38 \\ -104 & 40 \\ -14 \end{bmatrix}f(r,a+1)\right)$$ The matrices are shorthand for polynomials whose $r^ia^j$ coefficient is $a_{ij}$, using 0-indexing. With the help of the two smaller recurrences we can verify $$(r+1)^3\binom ra^2f(r,a)-\frac14(2r+3)(3r^2+9r+7)\binom{r+1}a^2f(r+1,a)+\frac1{16}(r+2)^3\binom{r+2}a^2f(r+2,a)=h(r,a+1)-h(r,a)$$ which telescopes to $$(r+1)^3A(r)-\frac14(2r+3)(3r^2+9r+7)A(r+1)+\frac1{16}(r+2)^3A(r+2)=0$$ But $B(r)$ satisfies the same recurrence, and the first two terms of $A(r)$ and $B(r)$ agree. This proves $A=B$.

Parcly Taxel
  • 105,904