Let $x_n$ be a sequence of positive integers defined by $x_n=9^n + 25^n -1$ for all $n \ge 2$
I conjectured that there exists at least one prime divisor of $x_n$ which contains $7 $ in its decimal representation for each $n$
Can anyone prove/disprove this?
This is not amenable to a heuristic argument. The prime factors are not normal for base ten. In particular, digits $\{0, 3, 5, 6, 8\}$ are avoided while $\{1, 4, 7, 9\}$ contend for most frequent (weighting by length). $8$ first occurs as a digit of $3274891$ for $n = 5$ but not in any factor for $n = \{9, 10, 11, 16, 21\}$. $0$ first occurs for $n = 8$, and then only because $152630937345$ happens to be fifteen times a prime number. Probably $9^n$ and $25^n$ are producing a subtle interference phenomenon in the p-adics.
For $n = 20$, $9$ is not a digit of a prime factor (the GPF has 18 digits) and $7$ occurs only as the units digit of three factors (out of six). This pattern repeats at $n = 24$, substituting $3$ for $7$ (the GPF has 22 digits and somehow excludes $\{3, 6, 9\}$ but not $0$, despite that $9^{24} + 25^{24} - 1 = 3552713678880267372432493847753985$ contains one $0$ and twice the expected number of $3$s).