# Consecutive numbers in rows of Pascal's triangle …

The fourteenth row of Pascal's triangle has an interesting property.

\begin{align} \binom{14}{4}+\binom{14}{5} &= 1001+2002 \\ =\binom{14}{6} &= 3003 \end{align}

This begs the follow-up question: are there other solutions $(n,k)$ to this system?

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

$$\binom{n}{k+2}=3\binom{n}{k}$$

More generally, are there other solutions to this equation?

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

Or equivalently,

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

Generalizing in a different way, are there other rows of Pascal's triangle which include arithmetic sequences of length $3$? I'd guess that there are no arithmetic sequences of length $4$, since the "second derivative" of $n$ choose $k$ with respect to $k$ is strictly negative.

• This is inspired by Will Jagy's observations here. I suspect that the other examples are very hard to find, and further that this question won't be answered easily, since it relates closely to Singmaster's conjecture (for which $3003$ is a notable numerical example). – Zubin Mukerjee Dec 28 '14 at 20:11
• oeis.org/A062730 – Edward Jiang Dec 28 '14 at 20:12
• @EdwardJiang Thanks! Why is this listed as the "formula" in the OEIS? I don't understand how it relates ... $$\frac{-5x^8+3x^7+7x^6-3x^5+5x^4-5x^3-12x^2+5x+7}{(1-x)(1-x^2)^2}$$ – Zubin Mukerjee Dec 28 '14 at 20:25
• That's a generating function. The sequence doesn't perfectly match up with your problem, though, so it shouldn't matter. Your problem calls for rows where three consecutive terms are in arithmetic progression, where the sequence just calls for rows where there exists three terms in arithmetic progression. – Edward Jiang Dec 28 '14 at 20:32
• @EdwardJiang Okay, thanks again. At least if people are looking for the consecutive terms in arithmetic sequence they only have to look at rows in the OEIS sequence you linked, so it narrows it down a lot ... – Zubin Mukerjee Dec 28 '14 at 20:34

## 1 Answer

Your questions have mostly already been answered in the comments; I compiled the answers from the various sources provided and added a computer search.

Your system

$$\binom{n}{k+1}=2\binom{n}{k}$$ $$\binom{n}{k+2}=3\binom{n}{k}$$

is a special case of an arithmetic progression of three consecutive binomial coefficients. In the answer to Generalized Case: Three Consecutive Binomial Coefficients in AP (linked to by hypergeometric in a comment), Jack shows that the triples of consecutive binomial coefficients in arithmetic progression form the family

$$\binom{a^2-2}{\binom a2-2},\binom{a^2-2}{\binom a2-1},\binom{a^2-2}{\binom a2}$$

with $a\ge3$ an integer. The ratio of the second member to the first is

$$\frac{a^2-\binom a2}{\binom a2-1}=\frac{a^2+a}{a^2-a-2}\;,$$

which takes the values $3,2,\frac53,\frac32$ for $a=3,4,5,6$, respectively. Thus your triple

$$\binom{14}4,\binom{14}5,\binom{14}6$$

is the only one that fulfils your system, but there is one more triple,

$$\binom71,\binom72,\binom73\;,$$

that fulfils the integer system

$$\binom{n}{k+1}=3\binom{n}{k}$$ $$\binom{n}{k+2}=5\binom{n}{k}$$

and one triple that at least has an integer ratio for the third member (and a half-integer ratio for the second):

$$\binom{34}{13},\binom{34}{14},\binom{35}{15}\;,$$

which fulfils the system

$$\binom{n}{k+1}=\frac32\binom{n}{k}\;,$$ $$\binom{n}{k+2}=2\binom{n}{k}\;.$$

This answers your first question. Your second question, whether there are further solutions to

$$\binom{n+1}{k+1}=\binom{n}{k+2}\;,$$

is answered in the Wikipedia article on Singmaster's conjecture that you linked to yourself in a comment: Singmaster himself proved in Repeated Binomial Coefficients and Fibonacci Numbers in $1975$ that there are infinitely many solutions, namely $n=F_{2i+2}F_{2i+3}-1$, $k=F_{2i}F_{2i+3}-1$ (where $F_n$ is the $n$-th Fibonacci number).

Your third question, about arithmetic sequences of length $3$ in rows of Pascal's triangle, has already been answered by Jack as regards consecutive entries. If we admit arbitrary, not necessarily consecutive entries, there seem to be no rigorous results, but there are interesting numerical ones. OEIS sequence A062730, which Edward linked to in a comment, states that up to the $1000$-th row, all arithmetic progressions except for one exception belong to two families. One family is of course the one described above; the other one is quite similar,

$$\binom{a^2-4}{\binom a2-4},\binom{a^2-4}{\binom a2-2},\binom{a^2-4}{\binom a2}\;,$$

with $a\ge4$ an integer. The exception is

$$\binom{19}4,\binom{19}6,\binom{19}7\;.$$

I did a computer search up to the $10000$-th row and found no further exceptions. (Here's the code.)

Regarding your fourth question about arithmetic sequences of length $4$, I agree with your negative answer, but the reason you give is not quite right. The "second derivative" of the binomial coefficient with respect to the lower index is not negative throughout; since with increasing upper index the binomial distribution increasingly resembles a Gaussian distribution, the "second derivative" changes from positive to negative and back to positive – in fact otherwise there couldn't be any arithmetic sequences of length $3$. Nevertheless, a given line crosses this bell-shaped curve at most three times, so there cannot be arithmetic sequences of length $4$ of consecutive entries. Nothing rigorous seems to be known about arithmetic sequences of length $4$ of not necessarily consecutive entries, but the computer search shows that there are none up to the $10000$-th row.