a(1) found by Patrick De Geest.
From Chai Wah Wu, Apr 18 2017: (Start)
Theorem: Let q be a term such that q = p_1+p_2+p_3 and p_1, p_2 p_3 are consecutive primes. Then q has 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. 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*10^k <= p_2 < 10^(k+1). Then by Bertrand's postulate, p_1 >= p_2/2 and p_3 <= 2*p_2. This implies that 10^(k+1) > p_1 >= 2*10^k, 4*10^k < p3 <= 2*10^(k+1) and thus 10^(k+1) < q < 4*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 shows that p_1 >= p_2*12/13 and p_3 <= p_2*14/13.
This shows that (2 12/13) p_2 <= q <= (3 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*10^k <= q <= (6 2/13) 10^k and thus q must start with the digit 3.
If p_2 start with the digit 3, then (8 10/13)10^k <= q <= (12 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 >= 2*10^k, i.e.
p_2 >= (1 11/13)10^k, p_1 >= (1 119/169) 10^k and q >= 4*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 >= 4* 10^k, i.e.
p_2 >= (3 9/13) 10^k, p_1 >= (3 69/169)10^k and q >= 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 7/19)10^k) and (3*10^k, (3 8/19)10^k) which amount to approximately 0.79*10^(k/2) palindromes to consider rather than a full search of 10*10^(k/2) palindromes of length k+1.
Conjecture: For all terms in, the corresponding primes p_1 has the same number of digits and the same first digit as p_2 and p_3.
(End)
