How many numbers in the interior of Pascal's triangle are Mersenne numbers?

Question

Consider the interior of Pascal's triangle, i.e. the triangle without numbers of the form $\binom{n}{0},\binom{n}{1},\binom{n}{n-1},\binom{n}{n}$.

How many numbers in the interior of Pascal's triangle are Mersenne numbers, that is, numbers of the form $2^n-1$ where $n\in\mathbb{Z^+}$?

I have found four such numbers: $\color{red}{\binom{6}{2}}=\color{red}{\binom{6}{4}}=15=2^4-1$ and $\color{red}{\binom{91}{2}}=\color{red}{\binom{91}{89}}=4095=2^{12}-1$. I searched, within the interior of Pascal's number, the smallest $10000$ numbers without repeats, the first $141$ rows, and $\binom{n}{2}$ up to $n=100000$.

Possibly related: There are no numbers of the form $2^n$ is the interior of Pascal's triangle (proof).

Context: I have been investigating Pascal's triangle, looking for new results. I thought of Mersenne primes, so this question naturally arose.

Answer 1

Partial results:

There are no further solutions to $\binom{n}{2} = 2^r-1$:

This is (one version of) the Ramanujan-Nagell equation, which has solutions only for $r = 0, 1, 2, 4, 12$. The proof is somewhat involved and requires some basic algebraic number theory.

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There are also no nontrivial solutions to $\binom{n}{3} = 2^r -1$:

We can rewrite the equation as

\begin{align*} n(n-1)(n-2) &= 6(2^r-1)\\ n^3-3n^2+2n+6 &= 6\cdot 2^r\\ (n+1)((n-2)^2+2) &= 6\cdot 2^r. \end{align*} The factor $(n-2)^2 + 2$ is never divisible by $4$; so the only way it can divide $6 \cdot 2^r$ is if it is equal to $2$, $3$, or $6$, corresponding to $n = 2, 3,$ and $4$.

Answer 2

This is a partial answer.

This answer proves the following three claims :

Claim 1 : $\binom n5=2^r-1$ has no non-trivial solutions.

Claim 2 : $\binom n7=2^r-1$ has no non-trivial solutions.

Claim 3 : $\binom n9=2^r-1$ has no non-trivial solutions.

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Let us consider $$\binom nm=2^r-1\tag1$$ where $m$ is odd.

The equation $(1)$ can be written as $$2^rm!=n(n-1)\cdots (n-m+1)+m!\tag2$$

We can see that if $m$ is odd, then the RHS is divisible by $n+1$.

Letting $N:=n+1$, $(2)$ can be written as $$2^rm!=(N-1)(N-2)\cdots (N-m)+m!\tag3$$ whose RHS is divisible by $N$ when $m$ is odd.

For $m=3$, Benjamin Wright has shown that there are no non-trivial solutions.

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Claim 1 : $\binom n5=2^r-1$ has no non-trivial solutions.

Proof :

For $m=5$, $(3)$ can be written as $$2^{r+3}\cdot 3\cdot 5=N\times f_5(N)$$ where $f_5(N)=N^4 - 15N^3+ 85 N^2 - 225 N + 274$.

Here, we have the followings :

• $N\ge 8$

• $N$ is not divisible by $4$
  (Proof : We have $f_5'(N)=(N - 5) (4(N-8)^2+39(N-8)+101)\gt 0$, so $f_5(N)\ge f_5(8)=330$. Suppose that $N$ is divisible by $4$. Then, $f_5(N)$ is of the form $2\times (\text{odd})$, so $f_5(N)\le 2\cdot 3\cdot 5=30$ which contradicts $f_5(N)\ge 330$)

• $N\not\equiv 3\pmod 4$
  (Proof : Suppose that $N\equiv 3\pmod 4$. Then, $n(n-1)\cdots (n-4)$ is divisible by $2^4$, so $\binom n5$ is even where $5!$ is not divisible by $2^4$)

• $N\not\equiv 1,2,3,4,5\pmod 8$
  (Proof : Suppose that $N\equiv 1,2,3,4,5\pmod 8$. Then, $n(n-1)\cdots (n-4)$ is divisible by $2^4$, so $\binom n5$ is even)

So, it is necessary that $N=30$, i.e. $n=29$ which is not sufficient.

Therefore, there are no non-trivial solutions.$\ \blacksquare$

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Claim 2 : $\binom n7=2^r-1$ has no non-trivial solutions.

Proof :

For $m=7$, $(3)$ can be written as $$2^{r+4}\cdot 3^2\cdot 5\cdot 7=N\times f_7(N)$$ where $f_7(N)=N^6 - 28 N^5 + 322 N^4 - 1960 N^3 + 6769 N^2 - 13132 N + 13068$.

Here, we have the followings :

• $N\ge 10$

• $N\not\equiv 0\pmod 9$
  (Proof : $f_7(N)$ is always divisible by $3$ since $f_7(N)\equiv N (N+1)^2 (N-1)^3\pmod 3$. Suppose that $N\equiv 0\pmod 9$. Then, $Nf_7(N)$ is divisible by $3^3$, a contradiction)

• $N$ is not divisible by $8$
  (Proof : We have $f_7'(N)=2 (N - 7) (3(N-10)^4+71(N-10)^3+631(N-10)^2+2487(N-10)+3708)\gt 0$, $f_7(N)\ge f_7(10)=18648$. Suppose that $N$ is divisible by $8$. Then, $f_7(N)$ is of the form $4\times (\text{odd})$, so $f_7(N)\le 2^2\cdot 3^2\cdot 5\cdot 7=1260$ which contradicts $f_7(N)\ge 18648$)

• $N\not\equiv 1,2,3\pmod 4$
  (Proof : Suppose that $N\equiv 1,2,3\pmod 4$. Then, $n(n-1)\cdots (n-6)$ is divisible by $2^5$, so $\binom n7$ is even where $7!$ is not divisible by $2^5$)

• $N\not\equiv 4\pmod 8$
  (Proof : Suppose that $N\equiv 4\pmod 8$. Then, $n(n-1)\cdots (n-6)$ is divisible by $2^5$, so $\binom n7$ is even)

Therefore, there are no non-trivial solutions.$\ \blacksquare$

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Claim 3 : $\binom n9=2^r-1$ has no non-trivial solutions.

Proof :

For $m=9$, $(3)$ can be written as $$2^{r+7}\cdot 3^4\cdot 5\cdot 7=N\times f_9(N)$$ where $f_9(N)=N^8 - 45 N^7 + 870 N^6 - 9450 N^5 + 63273 N^4 - 269325 N^3 + 723680 N^2 - 1172700 N + 1026576$.

Here, we have the followings :

• $N\ge 12$

• $N\not\equiv 0\pmod{27}$
  (Proof : $f_9(N)$ is always divisible by $9$ since $f_9(N)\equiv N^2 (N-1)^3(N+1)^3\pmod 9$. Suppose that $N\equiv 0\pmod{27}$. Then, $Nf_9(N)$ is divisible by $3^5$, a contradiction)

• $N$ is not divisible by $2^5$
  (Proof : We have $f_9'(N)=(N-9)(8(N-12)^6+333(N-12)^5+5733(N-12)^4+52191(N-12)^3+264999(N-12)^2+712116(N-12)+795580)\gt 0$, so $f_9(N)\ge f_9(12)=1693440$. Suppose that $N$ is divisible by $2^5$. Then, $f_9(N)$ is of the form $2^4\times (\text{odd})$, so $f_9(N)\le 2^4\cdot 3^4\cdot 5\cdot 7=45360$ which contradicts $f_9(N)\ge 1693440$)

• $N$ is even
  (Proof : Suppose that $N$ is odd. Then, since $n$ is even, $n(n-1)\cdots (n-8)$ is divisible by $2^8$, so $\binom n9$ is even where $9!$ is not divisible by $2^8$.)

• $N\not\equiv 2,4,6,8\pmod{16}$
  (Proof : Suppose that $N\equiv 2,4,6,8\pmod{16}$. Then, $n(n-1)\cdots (n-8)$ is divisible by $2^8$, so $\binom n9$ is even)

So, it is necessary that $N=12,14,16,28,30,42,48,60,80,90,112,126,140,144,240,252,336,560,720,1008,1260,1680,5040$.

So, it is necessary that $n=11,13,15,27,29,41,47,59,79,89,111,125,139,143,239,251,335,559,719,1007,1259,1679,5039$ none of which is sufficient.

Therefore, there are no non-trivial solutions.$\ \blacksquare$

Answer 3

Other partial results:

If one is willing to accept computer help on finding integer points on elliptic curves, then one can find all positive integers $n$, $r$ satisfying $\binom{n}{k}=2^r-1$ with $k=2$, $3$, $4$ or $6$.

In case $k=4$, on noting that $$\binom{n}{4}=\frac{(n^2-3n+1)^2-1}{24} , $$ the equation is rewritten $$(n^2-3n+1)^2+23=3\cdot 2^{r+3}.$$

For $r=3s$ for some positive integer $s$, this is equivalent to $$(3n^2-9n+3)^2=\left(3\cdot 2^{s+1}\right)^3-9\cdot 23.$$ Running the SAGE code

E = EllipticCurve([0,0,0,0,-207])

E.integral_points()

the computer returns $$(6 : 3 : 1), (12 : 39 : 1), (18 : 75 : 1), (31 : 172 : 1), (312 : 5511 : 1), (331 : 6022 : 1), (367806 : 223063347 : 1)].$$ Only the first two items from this list have the first entry of the required form $3\cdot 2^{s+1}$. Considering their second entries, one has to solve in integers $n>5$ the quadratic equations $n^2-3n+1= 1$ or $ 13$. As none of them is solvable under these conditions, our initial problem has no solution in this situation.

Assuming $r=3s+1$, one arrives at the elliptic equation $$(6n^2-18n+6)^2=\left(3\cdot 2^{s+2}\right)^3-36\cdot 23.$$ From the list $$[(12 : 30 : 1), (13 : 37 : 1), (24 : 114 : 1), (108 : 1122 : 1), (4464 : 298254 : 1)]$$ computed with SAGE we deduce $3\cdot 2^{s+2}=12$ and $n^2-3n+1= 5$ or $3\cdot 2^{s+2}=24$ and $n^2-3n+1= 19$. Hence, one gets $s=1$ and $n=6$, which is the known solution $\binom{6}{4}=2^4-1$.

Finally, when $r=3s+2$, then SAGE solves $$(12n^2-36n+12)^2=\left(3\cdot 2^{s+3}\right)^3-144\cdot 23$$ and finds unique integer positive solution $(16,28)$. Since the first entry does not have the form $3\cdot 2^{s+3}$, there is no $n>5$ and $r\equiv 2 \pmod 3$ with $\binom{n}{4}=2^r-1$.

Suppose $k=6$ and therefore $n>7$. As $$\binom{n}{6}=\frac{(n^2-5n)(n^2-5n+4)(n^2-5n+6)}{720},$$ we arrive at an equation of the type $$x(x+4)(x+6)+720=5\cdot 12^2\cdot 2^r.$$ Now we distinguish to subcases according to the parity of $r$.

When $r=2s$, the only integer points on the elliptic curve $$Y^2=X(X+20)(X+30)+720\cdot 5^3$$ having both coordinates positive are $(380, 7900),(870,26400), (3600,217500)$. None of the second entries has the required form $300\cdot 2^2$, so no solution for $\binom{n}{6}=2^{2s}-1$.

When $r=2s+1$, we get the elliptic curve $$Y^2=X(X+40)(X+60)+720\cdot 10^3.$$ Sieving the list returned by SAGE, it remains to solve the systems $(10(n^2-5n),300\cdot 2^s)=(60,1200)$ and $(10(n^2-5n),300\cdot 2^s)=(140,2400)$. None of them has a solution with $n>7$.

Clearly, the idea works for the already settled cases $k=2$, $3$.