Are there four consecutive binomial coefficients in a row in an arithmetic progression?

This is suggested by Will Jagy's comment to this question:

Find $n$ and $k$ if $\:\binom{n\:}{k-1}=2002\:\:\:\binom{n\:}{k}=3003\:\:$

Here is my answer: No. (If my algebra is correct - $P(error) > 1/e$ )

(Nope - I had an error pointed out by mathlove, below. Fortunately, the result is the same.)

To get 4 terms:

$\binom{n}{k} =a$, $\binom{n}{k+1} =a+d$, $\binom{n}{k+2} =a+2d$, $\binom{n}{k+3} =a+3d$.

From the first 3, $d =\binom{n}{k+2}-\binom{n}{k+1} =\binom{n}{k+1}-\binom{n}{k} $

or $\binom{n}{k+2}+\binom{n}{k} =2\binom{n}{k+1}\\ $

or $\frac{n!}{(k+2)!(n-k-2)!}+\frac{n!}{(k)!(n-k)!} =2\frac{n!}{(k+1)!(n-k-1)!} $

or, multiplying by $\frac{(k+2)!(n-k)!}{n!}$,

$(n-k-1)(n-k)+(k+1)(k+2) =2(k+2)(n-k) $

or $n^2-(2k+1)n+k(k+1)+k^2+3k+2 =2(kn+2n-k^2-2k) $

or $n^2-2kn-n+k^2+k+k^2+3k+2 =2kn+4n-2k^2-4k $

or $n^2-4kn+4k^2 =5n-8k-2 $

or $(n-2k)^2 =5n-8k-2 $.

The last 3 give the same thing but with $k+1$ for $k$, so that

$(n-2(k+1))^2 =5n-8(k+1)-2 $

or $(n-2k)^2-4(n-2k)+4 =5n-8k-10 $ or $(n-2k)^2 =9n-16k-14 $.

Equating, $5n-8k-2 =9n-16k-14 $ or $4n =8k+12 $ or $n =2k+3 $.

To summarize, if the terms are $p, q, r, s =a, a+d, a+2d, a+3d$ respectively, I have used $d =q-p = r-q =s-r $ to get $p+r=2q$ and $q+s=2r$.

Another relation would be $2a =p+2q-r $ and $a =q+r-s $ or $p+2q-r =2(q+r-s) $ or $p-3r+2s =0 $.

From this

$\begin{array}\\ 0 &=p-3r+2s\\ &=\binom{n}{k}-3\binom{n}{k+2}+2\binom{n}{k+3}\\ &=\frac{n!}{k!(n-k)!}-3\frac{n!}{(k+2)!(n-k-2)!}+2\frac{n!}{(k+3)!(n-k-3)!}\\ &=(k+1)(k+2)(k+3)-3(k+3)(n-k)(n-k-1)+2(n-k-2)(n-k-1)(n-k)\\ &\text{using Wolfy}\\ &=-4 k^3+12 k^2 n-12 k^2-9 k n^2+33 k n-2 k+2 n^3-15 n^2+13 n+6\\ &\text{again using Wolfy, substituting }n = 2k+3\\ &=-6 (k^2+5 k+6)\\ &=-6 (k+2)(k+3)\\ \end{array} $

Since none of the roots of this are a positive integer, there are no $n$ and $k$ for which the binomial coefficients are in arithmetic progression.

  • 1
    $\begingroup$ I would not have expected four in a row in any case, as the graph of the binomial coefficients along a fixed row mimics a scaled version of the Gaussian normal distribution, basically $e^{- x^2 / 2}.$ The middle term of three must be pretty near one of the inflection points of the Gaussian, at $x= \pm 1$ $\endgroup$ – Will Jagy Nov 18 '15 at 21:40
  • $\begingroup$ Jack D'Aurizio has a much simpler solution to the 3 in AP problem here: math.stackexchange.com/questions/1018687/… He shows that there are an infinite number of 3 term APs with $n=a^2-2, r=(a+1)(a-2)/2$. This can be readily used to show that there no 4 in AP. $\endgroup$ – marty cohen Nov 18 '15 at 21:46
  • 1
    $\begingroup$ So, in row 7, we get $7,21,35$ with difference $14.$ Never noticed $\endgroup$ – Will Jagy Nov 18 '15 at 21:48
  • $\begingroup$ Yep, and you can't write r+1 in that form also, so no solution. $\endgroup$ – marty cohen Nov 18 '15 at 21:50

After getting $$(n-k-1)(n-k)+(k+1)(k+2)=2(k+2)(n-k)$$ note that $$n^2+(-2k-1)n+k(k\color{red}{+}1)+\cdots$$

Then, $n=2k+3$ follows.

Now we have $$\binom{2k+3}{k},\binom{2k+3}{k+1},\binom{2k+3}{k+2},\binom{2k+3}{k+3}$$

This cannot be an arithmetic progression because we have $$\binom{2k+3}{k+1}-\binom{2k+3}{k}\gt 0$$ and $$\binom{2k+3}{k+3}-\binom{2k+3}{k+2}\lt 0.$$

  • $\begingroup$ Thanks. I have (I hope) fixed it. $\endgroup$ – marty cohen Nov 18 '15 at 21:37
  • $\begingroup$ That's very nice. Much easier than mine. $\endgroup$ – marty cohen Nov 18 '15 at 21:49

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