Let be $p_n$ a prime. Let $s\#$ denote the product of all primes up to $s$.
I ask for primes $p_n$ such that $p_n!$ has a number of digits which is a palindromic number and such that $n\#$ has as well a number of digits which is a palindromic number.
I found this example:
- $p_{1036}!=8263!$ has 28782 digits, which is a palindromic number; and $1036\#$ has 434 digits, which is as well a palindromic number.
Are there other primes $p$ of this type? Do you believe that they are infinitely many? Somebody could write a routine for finding solutions?
The following are the primes under $10^8$ that work:
$\begin{pmatrix} p_n & n & \text{digits in }p_n! & \text{digits in }n\# \\ 2 & 1 & 1 & 1 \\ 3 & 2 & 1 & 1 \\ 5 & 3 & 3 & 1 \\ 7 & 4 & 4 & 1 \\ 11 & 5 & 8 & 2 \\ 37 & 12 & 44 & 4 \\ 1777 & 275 & 5005 & 111 \\ 8263 & 1036 & 28782 & 434 \\ 10477 & 1282 & 37573 & 535 \\ 12641 & 1510 & 46364 & 636 \\ 19469 & 2208 & 75057 & 929 \\ 340601 & 29231 & 1736371 & 12621 \\ 681727 & 55195 & 3680863 & 23832 \\ \end{pmatrix}$