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As a web developer that programs in PhP, I enjoy running some scripts to see some of math wonders, however, PhP is limited to large calculations.

I wanted to see if there are any intersections for the following $2$ series:

A) $1 + 2 + 3+ 4 + 5 + \ldots$

B) $(1) + (1 + 2) + (1 + 2 + 3) +(1 + 2 + 3 + 4) + (1 + 2 + 3 + 4 + 5) + \ldots$

Out of the first $116410911$ positive integers, I have "found" the following "intersections" between A) and B):

At the number $10$:

$10 - (1+2+3+4)=0$

$10 - ((1)+(1+2)+(1+2+3))=0$

At the number $120$:

$120 - (1+2+3+4+5+6+7+8+9+10+11+12+13+14+15) = 0$

$120 - ((1)+(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)+(1+2+3+4+5+6+7)+(1+2+3+4+5+6+7+8)) = 0$

And I am not going to list the other procedures, but also:

At the numbers $1540$ and $7140$.

Can anyone confirm if there are any more "intersections"? In other words, am I to expect a finite case or an infinite case? Also, is it a coincidence that all four of them are numbers that are multiples of $10$?

soupless
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Isaac Brenig
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  • Related questions: https://math.stackexchange.com/questions/85442/are-there-surprisingly-identical-binomial-coefficients and https://math.stackexchange.com/questions/3200146/number-of-integral-solutions-to-elliptic-curve-binom-n2-binom-m3 – Gerry Myerson May 02 '21 at 07:00
  • Wikipedia reports that there are no more intersections between triangular and tetrahedral numbers. – Oscar Lanzi Mar 05 '23 at 20:08

2 Answers2

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These are the numbers that are both "triangular" ($n(n+1)/2$) and "tetrahedral" ($n(n+1)(n+2)/6$). They are tabulated at https://oeis.org/A027568. You have found all of them. There are several links to the literature at the OEIS site. It's unlikely that there is a wholly elementary proof, as elliptic curves are involved.

Gerry Myerson
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  • So that makes $120$ the only number that also intersects with $1⋅2 ⋅3 ⋅4 ⋅5$. Fascinating! I will try to understand the literature. – Isaac Brenig May 02 '21 at 07:19
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I think you are trying to find tuples $(a,b) $that satisfy the equation

$$\sum_{i = 1}^{a} i = \sum_{i = 1}^{b}\sum_{j =1}^{i}j$$

The left hand side is

$$\sum_{i = 1}^{a}i = \frac{a(a + 1)}{2}$$

and the right-hand side is

$$\sum_{i = 1}^{b}\sum_{j = 1}^{i}j = \frac{b(b + 1)(b + 2)}{6}.$$

Then, the equation becomes \begin{align*}\frac{a(a + 1)}{2} &= \frac{b(b + 1)(b + 2)}{6} \\ 3a(a + 1) &= b(b + 1)(b + 2)\end{align*}

This seems to be a Diophantine equation, but since I have no knowledge about it, someone could continue solving this.

To give a shot, I'll try to solve this by noticing that both sides have three factors, then we can equate the factors and see if it works.

We can let \begin{align*}3 &= b \\ a &= b + 1 \\ a + 1 &= b + 2.\end{align*} Notice that the second equation implies the third equation and $a = 4$ by solving for $a$ from the first equation. This is the first case that you gave which is for $10$.

The other cases is then left as an exercise.

By the way, I have no idea why you found tuples $(a,b)$ that are equal to a sum that is a multiple of 10. Someone might just answer this by continuing my solution.

soupless
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