Square pyramidal numbers and Tetrahedral numbers

In summary: So it's right at the edge of the internet, but out of reach for now.In summary, there are two types of numbers: square pyramidal numbers and tetrahedral numbers. The probability of finding a number that is both is very low, with only 1 being the solution to the equation. A proof for this fact was only discovered in 1988.
  • #1
Helios
269
63
There are square pyramidal numbers and tetrahedral numbers, defined

Square pyramidal numbers = n ( n + 1 )( 2 n + 1) / 6
1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, ...

Tetrahedral numbers = n ( n + 1 )( n + 2 ) / 6

1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, ...

and I was wondering if there's a number(s) besides 1 that is both.
 
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  • #2
I have to guess no. Purely for probabilistic reasons. For any number n, the probability that there is a pyramidal number equal to n(n+1)(n+2)/6 is ~1/n^2, falling off too fast as n -> infinity. So, for example, the probability of a hit for n>100 is ~0.01. Once we've checked the first 100 n's, we can be fairly sure that there won't be any hits beyond that.

But I have no idea how to give a proper proof.
 
  • #3
The answer is no. You are looking for solutions of the equation:

[tex]
\frac {n(n+1)(2n+1)}{6} = \frac {n(n+1)(n+2)}{6}
[/tex]

Cancelling terms we get

[tex]
2n+1 = n+2
[/tex]

Solving for n you find n = 1 as the only solution
 
  • #4
csco said:
The answer is no. You are looking for solutions of the equation:

[tex]
\frac {n(n+1)(2n+1)}{6} = \frac {n(n+1)(n+2)}{6}
[/tex]

Cancelling terms we get

[tex]
2n+1 = n+2
[/tex]

Solving for n you find n = 1 as the only solution

The OP is looking for solutions to

[tex]
\frac {n(n+1)(2n+1)}{6} = \frac {m(m+1)(m+2)}{6}
[/tex]
not

[tex]
\frac {n(n+1)(2n+1)}{6} = \frac {n(n+1)(n+2)}{6}
[/tex]
.
 
  • #5
According to Mathworld, 1 is the only solution, and this fact was only proven in 1988 (so, no easy proof is forthcoming).
 
  • #6
hamster143 said:
According to Mathworld, 1 is the only solution, and this fact was only proven in 1988 (so, no easy proof is forthcoming).

really? Can you give a cite?
 
  • #7
MathWorld gives
Beukers, F. "On Oranges and Integral Points on Certain Plane Cubic Curves." Nieuw Arch. Wisk. 6, 203-210, 1988.​
but Nieuw Archief voor Wiskunde's online archives only go back to 2000.
 

Related to Square pyramidal numbers and Tetrahedral numbers

1. What are square pyramidal numbers?

Square pyramidal numbers are numbers that can be represented as a square pyramid. This means that if you were to stack square blocks on top of each other, with each layer having one more block than the layer below it, the total number of blocks in the pyramid would be a square number. For example, 1, 5, 14, and 30 are all square pyramidal numbers.

2. How do you find the nth square pyramidal number?

The formula for finding the nth square pyramidal number is n(n+1)(2n+1)/6. For example, the 5th square pyramidal number would be 5(5+1)(2(5)+1)/6 = 30.

3. What are tetrahedral numbers?

Tetrahedral numbers are numbers that can be represented as a tetrahedron. This means that if you were to stack triangular blocks on top of each other, with each layer having one more block than the layer below it, the total number of blocks in the tetrahedron would be a triangular number. For example, 1, 4, 10, and 20 are all tetrahedral numbers.

4. How do you find the nth tetrahedral number?

The formula for finding the nth tetrahedral number is n(n+1)(n+2)/6. For example, the 4th tetrahedral number would be 4(4+1)(4+2)/6 = 20.

5. What is the relationship between square pyramidal numbers and tetrahedral numbers?

The nth square pyramidal number is equal to the sum of the first n tetrahedral numbers. In other words, the sum of the first n tetrahedral numbers is equal to the nth square pyramidal number. This relationship can be represented by the formula n(n+1)(n+2)/6 = (1/6)(n)(n+1)(n+2)(n+3)/4, where n represents the nth number in the sequence.

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