$\binom{n}{k} : \binom{n}{k+1} : \binom{n}{k+2} = a : b : c$

It is a rather surprising fact (to me, at least) that $\displaystyle \binom{14}{4} = 1001$; $\displaystyle \binom{14}{5} = 2002$; $\displaystyle \binom{14}{6} = 3003$.

Actually, this is the only instance where three consecutive binomial coefficients are in the ratio $\displaystyle 1 : 2 : 3$. I found it quite interesting to investigate under what conditions consecutive members of a row of Pascal’s triangle can be in the ratio $\displaystyle a : b : c$, where $\displaystyle a,b,c$ are positive integers with $\displaystyle \mathrm{gcd}(a,b,c) = 1$ and $\displaystyle a < b < c$, except where otherwise stated. That is, only the left-hand side of the triangle will be considered.

$$\binom{n}{k} : \binom{n}{k+1} : \binom{n}{k+2} = a : b : c$$

Cancelling and rearranging,

$\displaystyle b(k + 1) = a(n - k)$......….....[1]
$\displaystyle c(k + 2) = b(n - k - 1)$.........[2]

$$n = \frac{a(b + c) + c(a + b)}{(b^2 - ac)}$$

$$k = \frac{a(b + c)}{b^2 - ac} - 1$$

Therefore n and k are integers iff $\displaystyle b^2-ac$ divides both $\displaystyle a(b + c)$ and $\displaystyle c(a + b)$.

$\displaystyle n > 0$ implies $\displaystyle b^2 > ac$

$\displaystyle k\ge 0$ implies $\displaystyle c \ge \frac{b(b - a)}{2a}$

Hence a third condition is $\displaystyle \frac{b^2}{a} > c \ge \frac{b(b - a)}{2a}$

Perhaps the most interesting special case is $\displaystyle c = a + b$, when for

$\displaystyle a,b < 100$ there are only five solutions. Namely,

$\displaystyle (a,b,c) = (1,2,3)$ gives $\displaystyle \{n,k\} = \{14,4\}$

$\displaystyle (a,b,c) = (3,5,8)$ gives $\displaystyle \{n,k\} = \{103,38\}$

$\displaystyle (a,b,c) = (8,13,21)$ gives $\displaystyle \{n,k\} = \{713,271\}$

$\displaystyle (a,b,c) = (21,34,55)$ gives $\displaystyle \{n,k\} = \{4894,1868\}$

$\displaystyle (a,b,c) = (55,89,144)$ gives $\displaystyle \{n,k\} = \{33551,12814\}$

That is, there are solutions only when $\displaystyle (a,b,c) = (F(2m), F(2m + 1), F(2m + 2))$

$\displaystyle m = 1,2,3...,$ where $\displaystyle F(m)$ is the $\displaystyle m^{th}$ Fibonacci number.

Generally,

$\displaystyle \{n,k\} = \{F(2m + 2)F(2m + 3) - 1, F(2m)F(2m + 3) - 1\}$

All solutions satisfy $\displaystyle F(2m)n = F(2m+2)k + F(2m+1)$

Where possible I have been able to derive formulae for $\displaystyle \{n,k\}$ for all special cases I could think of (eg. $\displaystyle a,b,c$ in arithmetic progression) except for the case $\displaystyle c = a^2$.

For $\displaystyle a,b < 3000$ there are only three solutions:

$\displaystyle (a,b,c) = (1,2,1)$ gives $\displaystyle \{n,k\} = \{2,0\}$

$\displaystyle (a,b,c) = (2,3,4)$ gives $\displaystyle \{n,k\} = \{34,13\}$

$\displaystyle (a,b,c) = (13,47,169)$ gives $\displaystyle \{n,k\} = \{1079,233\}$

Letting $\displaystyle c = a^2$ and dividing equation [2] by [1] leads to $\displaystyle a(k + 1)(k + 2) = (n - k)(n - k - 1)$

Rearranging, all solutions satisfy $\displaystyle n^2 - (2k + 1)n - (k + 1)[(a - 1)k + 2a] = 0$

The discriminant $\displaystyle D$ of the above quadratic is $\displaystyle 4a(k + 1)(k + 2) + 1$, and a necessary and sufficient condition for $\displaystyle n$ to be an integer is that this expression be a perfect square.

$\displaystyle a = 1, k = 0$ gives $\displaystyle D = 9 = 3^2$

$\displaystyle a = 2, k = 13$ gives $\displaystyle D = 1681 = 41^2$

$\displaystyle a = 13, k = 233$ gives $\displaystyle D = 2859481 = 1691^2$

And my question is: can one prove (or disprove) that there are no more solutions?

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What's the question, exactly? –  Hans Lundmark Oct 30 '10 at 13:04
The question is 'Can we prove (or disprove) that there are no more solutions for the case c = a^2?' –  ThudanBlunder Oct 30 '10 at 15:38
Yes we can prove it. See "Which way did the bicycle go" page 184. –  Jaska Oct 30 '10 at 16:39
Jaska, I don't have that book, and page 184 is not available at Google Books. What exactly does it solve and in connection with what? –  ThudanBlunder Oct 30 '10 at 17:33
@ThudanBlunder: You can see it on Google Books if you search for "binomial". –  Hans Lundmark Oct 30 '10 at 18:04

Suppose $\binom{n}{k},\binom{n}{k+1},\binom{n}{k+2}$ are in ratio $1:2:3$. Then $\frac{n-k}{k+1}=2$ and $\frac{n-k-1}{k+2}=\frac{3}{2}$. Solving these two equations gives $n=14$ and $k=4$.