Smith Numbers are Infinite in Number/Lemma

Theorem
Let $\map S m$ denote the sum of the digits of a positive integer $m$.

Let $\map {S_p} m$ denote the sum of the digits of the prime decomposition of $m$.

Let $\map N m$ denote the number of digits in $m$.

Suppose $m = p_1 p_2 \dots p_r$, where $p_i$ are prime numbers, not necessarily distinct.

Then $\map {S_p} m < 9 \map N m - 0.54 r$.

Proof
Let $b_i = \map N {p_i} - 1, i = 1, 2, \dots, r$.

Let $b = b_1 + b_2 + \dots + b_r$.

Since a prime cannot be a multiple of $9$, $\map S {p_i}$ cannot be a multiple of $9$ either.

Hence:
 * $\map S {p_i} \le 9 \map N {p_i} - 1 = 9 b_i + 8$

We partition the $p_i$ into $9$ disjoint classes by means of the following:

Define $c_i = \map S {p_i} - 9 b_i$.

By the above, $c_i \le 8$ and $c_i \ne 0$.

Let $n_0$ be the number of $i$ such that $c_i$ is (strictly) negative, and $n_j$ be the number of $i$ such that $c_i = j$ for $1 \le j \le 8$.

Then:

where the last sum is over the integers $i$ where $c_i$ is (strictly) negative.

There are $n_0$ of these integers, so we also have:

We have $\map S {p_i} = 9 b_i + c_i$.

For $c_i < 0$ we will use the inequality:
 * $p_i > 10^{\map N {p_i} - 1} = 10^{b_i}$

For $1 \le c_i \le 8$ we would have:
 * $p_i \ge \paren {c_i + 1} \times 10^{b_i} - 1$

since this number on the right is the smallest integer $N$ with $\map S N = 9 b_i + c_i$.

We also have:
 * $\paren {c_i + 1} \times 10^{b_i} - 1 = \paren {c_i + 0.9} \times 10^{b_i} + 10^{b_i - 1} - 1 \ge \paren {c_i + 0.9} \times 10^{b_i}$

for $b_i > 0$ (i.e. when $p_i \ne 2, 3, 5, 7$).

Finally for $b_i = 0$, we have:
 * $p_i = \map S {p_i} = c_i = c_i 10^{b_i}$

It follows that:

Write $m = a \times 10^{\map N m - 1}$ where $1 \le a < 10$.

By taking logarithm (base $10$) on both sides, we get:

Combining $(1)$ and $(2)$, we have: