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Representations as $p + 2^x + 11 \cdot 2^y$ with $p \equiv 1 \pmod 6$

Number of ways to write the $n$-th positive odd integer in the form $p + 2^x + 11 \cdot 2^y$ with $p$ a prime congruent to $1 \bmod 6$ and $x, y$ positive integers. $$a(n) = \left|\left{(p, x, y) : p + 2^x + 11 \cdot 2^y = 2n - 1 \text{ with } p \text{ prime}, p \equiv 1 \pmod 6, x, y \in \mathbb{Z}^+\right}\right|.$$

References:

namespace OeisA157237

Number of representations of $2n - 1$ as $p + 2^x + 11 \cdot 2^y$ with $p \equiv 1 \pmod 6$.

def a (n : ) : := if n = 0 then 0 else let N := 2 * n - 1 x Finset.Icc 1 N, y Finset.Icc 1 N, if 2 ^ x + 11 * 2 ^ y < N (N - (2 ^ x + 11 * 2 ^ y)).Prime (N - (2 ^ x + 11 * 2 ^ y)) % 6 = 1 then 1 else 0

Value of the sequence a at 1.

@[category test, AMS 11] theorem a_1 : a 1 = 0 := a 1 = 0 All goals completed! 🐙

Value of the sequence a at 2.

@[category test, AMS 11] theorem a_2 : a 2 = 0 := a 2 = 0 All goals completed! 🐙

Value of the sequence a at 16.

@[category test, AMS 11] theorem a_16 : a 16 = 1 := a 16 = 1 All goals completed! 🐙

Value of the sequence a at 17.

@[category test, AMS 11] theorem a_17 : a 17 = 1 := a 17 = 1 All goals completed! 🐙

Value of the sequence a at 18.

@[category test, AMS 11] theorem a_18 : a 18 = 0 := a 18 = 0 All goals completed! 🐙

Value of the sequence a at 19.

@[category test, AMS 11] theorem a_19 : a 19 = 2 := a 19 = 2 All goals completed! 🐙

On Feb. 24, 2009, Zhi-Wei Sun conjectured that $a(n) = 0$ if and only if $n < 16$ or $n \in {18, 21, 24, 51, 84, 1011, 59586}$; in other words, except for $35, 41, 47, 101, 167, 2021, 119171$, any odd integer greater than $30$ can be written as the sum of a prime congruent to $1 \bmod 6$, a positive power of $2$ and eleven times a positive power of $2$.

@[category research open, AMS 11] theorem conjecture (n : ) (hn : 0 < n) : a n = 0 n 15 n = 18 n = 21 n = 24 n = 51 n = 84 n = 1011 n = 59586 := n:hn:0 < na n = 0 n 15 n = 18 n = 21 n = 24 n = 51 n = 84 n = 1011 n = 59586 All goals completed! 🐙end OeisA157237