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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?

  • 4
    What's the question, exactly?2010-10-30
  • 1
    The question is 'Can we prove (or disprove) that there are no more solutions for the case c = a^2?'2010-10-30
  • 1
    Yes we can prove it. See "Which way did the bicycle go" page 184.2010-10-30
  • 0
    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?2010-10-30
  • 0
    @ThudanBlunder: You can see it on Google Books if you search for "binomial".2010-10-30
  • 0
    ...although all that I see there is the 1:2:3 case, which is not what you asked about if I understand correctly (I have to admit that I haven't read your question in detail).2010-10-30
  • 0
    Sorry. I misread the question. There has been dealt only the case 1:2:3.2010-10-30
  • 0
    Google Books shows different results (and different pages available for preview) when you search from different geographical regions. Even books that are "Full view" in one country may be "No preview available" in another.2010-11-09

1 Answers 1

2

Sketch:

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 $k=4$.

  • 0
    Janka, what question have you answered there?2010-10-30
  • 0
    I apologize. I misread the question and thought an author want to find all binomial coefficients on the same row on the Pascal triangle which are on ratio 1:2:3.2010-10-31