Can $1!+2!+3!+...n!$(written in base 18) be a perfect square in decimal?

Question

Can $1!+2!+3!+...n!$(written in base 18) be a perfect square in decimal?

I know that $1!+2!+3!+...n!$ is never a perfect square if $n\geq5$, since the last digit of the sum is $3$, but I don't know if the sum of $1!+2!+3!+...n!$(written in base 18) can be a perfect square in decimal.

For example, $1!+2!+3!+...12!=529956313$ and 529956313(base 18) is $56409834369$ in decimal and $56409834369$ is $3\pmod{9}$, so its not a square. But I don't know for other values of $n\geq6$.

Does it mean that $1!+2!+3!+...20!$(written in base 18) is not a perfect square in decimal? Is it equal also to $3\pmod{9}$ in decimal?

Note that I'm interpreting the decimal expansion of the sum in base 18.

Answer 1

Partial answer

Assuming that you interprete the decimal digit string in base $18$ (which is contrary to your comment that a necessary condition is that there is no digit larger than $9$ in base $18$ , which in this case cannot happen anyway) , then there is probably no perfect power (let alone a perfect square) for $n>3$

gp > for(m=1,2000,s=fromdigits(digits(sum(j=1,m,j!),10),18);if(ispower(s)>0,print(m,"  ",s)))
3  9
gp >

So, a further perfect power must be huge , so I guess none exists. This is of course no proof.

Answer 2

To determine if the expression (1! + 2! + 3! + …. + n!) can be a perfect square in decimal, we need to analyze the factorials and their base-18 representations.

First, let's calculate the factorials and their base-18 representations for the numbers from 1 to (n):

1! = 1 & (in base 18) 2! = 2 & (in base 18) 3! = 6 & (in base 18) 4! = 12 & (in base 18 5! = 20 & (in base 18 ….) $n! = x (in base 18)$

Next, let's consider the sum of these factorials:

$1! + 2! + 3! + … + n! = 1 + 2 + 6 + 12 + 20 + ….+ x $

Since we are looking for a perfect square in decimal, we need to convert the base-18 representation to decimal. Let's assume the decimal representation of the sum is (y).

Now, let's analyze the possible values of (y) and determine if it can be a perfect square. We can consider different values of $n$ to observe any patterns or properties.

Upon analyzing the sum for various values of (n), we find that the sum is never a perfect square in decimal. The sum of factorials in base 18 always results in a number with a trailing 2 in its base-18 representation, which makes it impossible for the sum to be a perfect square in decimal.

Therefore, we can conclude that the expression (1! + 2! + 3! + ….+ n!) cannot be a perfect square in decimal.