**Theorem**: All terms of A113846 have an odd number of digits and the first digit (and last digit) is either 3 or 9. In addition, $p_2$ and $p_3$ have the same first digit and $p_2$ and $p_3$ have the same number of digits as $q$. The first digit of $p_2$ (and $p_3$) is equal to the first digit of $q$ divided by 3, i.e. if the first digit of $q$ is $3$, then the first digit of $p_2$ and $p_3$ is $1$.

**Proof**: First note that all palindromes with an even number of digits is divisible by 11. Let $p_1$, $p_2$, $p_3$ be 3 consecutive primes such that $p_2$ is a palindrome and $q = p_1+p_2+p_3$ is a palindromic prime. It is clear that $q, p_2 > 11$ and have a odd number of digits since the first term of A113846 is 31513.

The first digit of $p_2$ cannot be 2 since $p_2$ is odd. Next we show that the first digit of $p_2$ is either 1 or 3.

Let $k$ be an even integer such that $p_2$ has $k+1$ digits. Suppose the first digit of $p_2$ is 4 or larger, i.e. $4\cdot 10^k \leq p_2 < 10^{k+1}$. Then by Bertrand's postulate (or Bertrand-Chebyshev theorem), $p_1\geq \frac{p_2}{2}$ and $p_3 \leq 2p_2$. This implies that $10^{k+1} > p_1 \geq 2\cdot 10^k$, $4\cdot 10^k < p3 \leq 2\cdot 10^{k+1}$ and thus $10^{k+1} < q < 4\cdot 10^{k+1}$, i.e. $q$ has an even number of digits, a contradiction.

Next we show that the first digit of $q$ is 3 or 9. To do this, we need a stronger result than Bertrand's postulate for the prime gap. Since $p_2 > 647$, Rohrback and Weis's 1964 result [1] shows that $p_1 \geq \frac{12}{13}p_2$ and $p_3 \leq \frac{14}{13}p_2$.

This shows that $2\frac{12}{13} p_2 \leq q \leq 3\frac{1}{13} p_2$. Since $q$ is prime, it cannot start with the digit 5. If $p_2$ start with the digit 1, then $2\cdot 10^k \leq q \leq 6 \frac{2}{13} 10^k$ and thus $q$ must start with the digit 3.

If $p_2$ start with the digit 3, then $8\frac{10}{13}10^k \leq q \leq 12\frac{4}{13}10^k$. Since $q < 10^{k+1}$, this means that $q$ must start with the digit 9.

Clearly $p_3$ must have the same number of digits as $p_2$ and as $q$.

Finally, we show that $p_3$ has the same starting digit as $p_2$. Suppose $p_2$ starts with the digit 1. If $p_3$ has a different starting digit, this must mean that $p_3 \geq 2\cdot 10^k$, i.e.

$p_2 \geq 1\frac{11}{13}10^k$, $p_1\geq 1\frac{119}{169} 10^k$ and $q \geq 4\cdot 10^k$, a contradiction.

Suppose $p_2$ start with the digit 3. If $p_3$ has a different starting digit, this must mean that $p_3 \geq 4\cdot 10^k$, i.e.

$p_2 \geq 3\frac{9}{13} 10^k$, $p_1 \geq 3\frac{69}{169}10^k$ and $q\geq 10^{k+1}$, a contradiction. QED

This argument also shows that to search for terms, for each $k$, one only needs to consider for $p_2$ the palindromic primes in the ranges $(10^k, 1\frac{7}{19}10^k)$ and $(3\cdot 10^k, 3 \frac{8}{19}\cdot 10^k)$ which amount to approximately $0.79\cdot 10^{\frac{k}{2}}$ palindromes to consider rather than a full search of $10\cdot 10^{\frac{k}{2}}$ palindromes of length $k+1$.

**Conjecture**: For all terms in A113846, the corresponding primes $p_1$ has the same number of digits and the same first digit as $p_2$ and $p_3$.

**References:**

1. H. Rohrback and J. Weis, "Zum finiten Fall des Bertrandschen Postulats", J. Reine Angew. Math., 214/215 (1964), 432–440.

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