Is the hexagonal cannonball number the only one that ends with $6$?

Question

I've been searching for cannonball numbers, namely polygonal numbers that are also pyramidal numbers with the same number of sides, patterned after the famous cannonball number $4900$, the square that is also a square pyramidal number. In my initial search I found many numbers that end in $0$, $1$, or $5$, but only one that ends in $6$: $946$, the hexagonal cannonball number which is a hexagon with a side length of $22$ or a hexagonal pyramid $11$ levels high. Is that the only such number?

Answer 1

The answer is no according to this page which says

(1) $101337426$ is a $145$-agon cannonball number.

(2) $1169686$ is a $322$-agon cannonball number.

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The claim $(1)$ is true since $$\frac{(s-2)m^2-(s-4)m}{2}=\frac{(s-2)n^3+3n^2-(s-5)n}{6}=101337426$$ holds for $(s,m,n)=(145,1191,162)$.

The claim $(2)$ is true since $$\frac{(s-2)m^2-(s-4)m}{2}=\frac{(s-2)n^3+3n^2-(s-5)n}{6}=1169686$$ holds for $(s,m,n)=(322,86,28)$.

Answer 2

I know I'm a little bit late to the chat, but I have done an extensive search through the cannonball numbers and I've found 10 such numbers so far:

(1) The aforementioned 6-agon cannonball number

(2) A 10-agon cannonball number: 368050005576

(3) An 11-agon cannonball number: 7248070597636

(4) A 17-agon cannonball number: 1580765544996

(5,6) The 145 and 322-agons mentioned in the reply by mathlove

(7) A 9325-agon cannonball number: 3176083959788026

(8) A 16420-agon cannonball number: 913053565546276

(9) A 19605-agon cannonball number: 5519583702676

(10) A 125070-agon cannonball number: 890348736143873526

Your post has inspired me to look at my list again, and I've managed to pull out an interesting pattern from the data: While there are clearly quite a few of these numbers that end with 6, all of the cannonball numbers I found (searching as high as a million sides and as high as 2^53) seem to end with either 0, 1, 5, or 6. After doing a bit more research I found that every number of sides congruent to 2 mod(3) that's greater than 5 has at least one solution that can be found with a formula:

$C_s=\frac{1}{162}\left(s^{7}-14s^{6}+66s^{5}-91s^{4}-133s^{3}+309s^{2}+70s-38\right)$

(http://oeis.org/A027696 provides the numbers of sides which don't follow this pattern). All of these seem to only end with the digits 1 or 5. This might be provable with induction, but I have no clue how to prove anything for solutions that don't follow this pattern.

Edit: Just proved that all of the numbers following that polynomial end in 1 or 5 using induction. It's a bit of a pain, and you have to split it into 10 individual cases, but it's definitely doable with a bit of patience and Wolfram Alpha to help expand the polynomials