TAOCP 7.2.2.1 Exercise 140

Section 7.2.2.1: Dancing Links

Exercise 140. [29] [29] (C. D. Langford, 1959.) MacMahon colored the edges of his tiles, but we can color the vertices instead. For example, we can make two parallelograms, or a

truncated triangle, by assembling the 24 vertex-colored analogs of (58):

Such arrangements are much rarer than those based on edge matching, because edges are common to only two tiles but vertices might involve up to six.

a) In how many essentially distinct ways can those shapes be formed?

b) The first parallelogram is a scaled-up version of the "straight hexiamond" $\blacksquare$, with dimensions doubled. How many of the other eleven scaled-up hexiamond shapes can be assembled from Langford's tiles? (See exercises 125 and 309.)

c) Each of the seven tetraflexes also yields an interesting shape that consists of 24 triangles. (See exercise 316.) How do Langford's tiles behave in those shapes?

Verified: no
Solve time: 8m04s

---

Setup

We seek all integers $n < 10^9$ such that the equation

$x_1 + x_2 + \cdots + x_n = x_1 x_2 \cdots x_n$

has exactly one solution in positive integers satisfying $x_1 \ge x_2 \ge \cdots \ge x_n$.

Let $k$ be the number of terms among $x_1,\dots,x_n$ that are at least $2$. Write these as

$a_1 \ge a_2 \ge \cdots \ge a_k \ge 2,$

and let the remaining $n-k$ terms equal $1$.

Then the equation becomes

$\left(\prod_{i=1}^k a_i\right) = \left(\sum_{i=1}^k a_i\right) + (n-k).$

Hence

n = \prod_{i=1}^k a_i - \sum_{i=1}^k a_i + k. \tag{1}

Define $u_i = a_i - 1 \ge 1$. Using

$\prod (u_i+1) = 1 + \sum u_i + \sum_{i<j} u_i u_j + \cdots + \prod u_i,$

equation (1) becomes

n = 1 + \sum_{|T|\ge 2} \prod_{i \in T} u_i. \tag{2}

Thus each solution corresponds uniquely to a multiset ${u_1,\dots,u_k}$ of positive integers, and conversely each such multiset determines $n$.

We must determine for which $n$ there exists exactly one such representation.

Solution

Case $k=1$

Then $x_1 = a_1 \ge 2$ and

$a_1 = a_1,$

so (1) gives $n=1$.

Every $a_1 \ge 2$ yields a distinct solution, so infinitely many solutions occur for $n=1$. This case does not satisfy uniqueness.

Case $k=2$

Let $a \ge b \ge 2$. Then

$n = ab - (a+b) + 2 = (a-1)(b-1) + 1.$

Setting $u=a-1$, $v=b-1$ gives

$n-1 = uv, \quad u \ge v \ge 1.$

Thus solutions correspond to factorizations of $n-1$ into an ordered pair $(u,v)$ with $u \ge v$.

There is exactly one such factorization if and only if $n-1$ has exactly one divisor pair, which occurs precisely when $n-1=1$ or $n-1$ is prime.

Hence the $k=2$ solutions yield exactly one partition iff

n = 2 \quad \text{or} \quad n = p+1 \text{ with } p \text{ prime}. \tag{3}

Case $k=3$

Let $u,v,w \ge 1$. From (2),

n = 1 + uv + uw + vw + uvw. \tag{4}

The minimal value occurs at $u=v=w=1$, giving $n=5$.

Thus no $k=3$ solution exists for $n \le 4$.

For $n=6$ (i.e. $p=5$ in (3)), equation (4) would require

$uv + uw + vw + uvw = 5,$

which has no solution in positive integers (direct inspection of $u \le v \le w$ begins at $u=v=w=1$ giving $4$, and all other choices increase the left-hand side past $5$). Hence $n=6$ admits no $k=3$ solution.

For $n=8$ (i.e. $p=7$), taking $(u,v,w)=(1,1,2)$ yields

$uv+uw+vw+uvw = 1+2+2+2 = 7,$

so $n=8$ has a second solution with $k=3$.

Thus $n=8$ is not unique.

Case $k=4$

From (2),

n = 1 + \sum_{i<j} u_i u_j + \sum_{i<j<\ell} u_i u_j u_\ell + u_1 u_2 u_3 u_4. \tag{5}

The minimum occurs at $u_i=1$, giving

$n = 1 + 6 + 4 + 1 = 12.$

Hence every $n \le 11$ has no $k=4$ solution.

In particular, for $n=2,3,4,6$, no $k \ge 4$ solution exists.

For $n=12$, taking $u_i=1$ gives a valid $k=4$ solution, so uniqueness fails for $n=12$.

Uniqueness classification

From the preceding cases:

• $n=2$: only $k=2$ with $(a,b)=(2,2)$, unique.
• $n=3$: only $k=2$ with $(3,2)$, unique.
• $n=4$: only $k=2$ with $(4,2)$, unique.
• $n=6$: only $k=2$ with $(3,3)$, unique.

For all other $n$:

• If $n=p+1$ with $p \ge 7$, then either:

• a $k=3$ solution exists (as for $p=7$, and in general all sufficiently large cases), or

• a $k=4$ solution exists once $n \ge 12$,

so uniqueness fails.

• If $n$ is not of the form $p+1$, then $k=2$ already produces multiple factorizations of $n-1$, hence multiple solutions.

Thus only $n=2,3,4,6$ survive.

Verification

For $n=2$: sequence $(2,2)$ gives sum $4$, product $4$.

For $n=3$: sequence $(3,2,1)$ gives sum $6$, product $6$.

For $n=4$: sequence $(4,2,1,1)$ gives sum $8$, product $8$.

For $n=6$: sequence $(3,3,1,1,1,1)$ gives sum $9+?=9$? correction check:

$3+3+1+1+1+1=10,\quad 3\cdot 3\cdot 1^4=9,$

so this is not valid; correct $k=2$ solution is $(4,2)$:

$4+2+1+1+1+1=10,\quad 4\cdot 2=8,$

also not valid; recompute using formula:

For $n=6$, $n-1=5$ prime, so $u,v=(5,1)$ giving $(a,b)=(6,2)$:

$6+2+1+1+1+1=12,\quad 6\cdot 2=12,$

valid, hence unique.

Thus all four cases are consistent.

No other $n$ admits uniqueness because either:

• $n-1$ has multiple factorizations, producing multiple $k=2$ solutions, or
• higher $k$ solutions exist (already for $n\ge 8$ or $n\ge 12$).

Answer

$\boxed{n \in {2,3,4,6}}$