Is $k=11$ the largest value of $k$ such that $(\lfloor\sqrt{1!+2!+3!+\cdots+k!}\rfloor+1)^{2} - (1!+2!+3!+\cdots+k!)$ is a perfect power?
While I playing on my calculator, I observed the following:
$(\lfloor\sqrt{1!+2!+3!+4!+5!}\rfloor+1)^{2} - (1!+2!+3!+4!+5!)=16=2^{4}$
$(\lfloor\sqrt{1!+2!+3!+\cdots+6!}\rfloor+1)^{2} - (1!+2!+3!+\cdots+6!)=27=3^{3}$
$(\lfloor\sqrt{1!+2!+3!+\cdots+7!}\rfloor+1)^{2} - (1!+2!+3!+\cdots+7!)=16=2^{4}$
$(\lfloor\sqrt{1!+2!+3!+\cdots+10!}\rfloor+1)^{2} - (1!+2!+3!+\cdots+10!)=2187=3^{7}$ (This is also true when $k=11$.)
But is $11$ the largest possible value?
I have no idea, however I know the following:
- I know that the last digit of $1!+2!+3!+\cdots+k!$ is $3$ when $k\geq4$ and the last digit of $(\lfloor\sqrt{1!+2!+3!+\cdots+k!}\rfloor+1)^{2}$ never ends with $5$, so if we want for $(\lfloor\sqrt{1!+2!+3!+\cdots+10!}\rfloor+1)^{2}- (1!+2!+3!+\cdots+k!)$ to be a perfect square, then the last digit of equation must either $4$ or $9$.
I’am correct?
I did my best by finding all values of $k\leq10^{4}$ using Pari GP, but I was not successful.