The sequence $a(n)=(2^n \bmod p)^{p-1} \bmod p^2$

Question

Related to this question.

Let $p$ be prime and $n$ positive integer.

Define $a(n)=(2^n \bmod p)^{p-1} \bmod p^2$

Let $D(n)$ be the base $2$ discrete logarithm of $a(n)$, i.e. given $p,a(n)$ we have $2^{D(n)} \bmod p^2=a(n)$. We can efficiently compute $D(n)=k(p-1)$ via p-adic logarithms and code is given in the linked question.

Let $D(n)$ be computed via p-adic algorithms, probably differing from the smallest logarithm by a small factor.

Strong numerical evidence suggests that three consecutive $D(i)$ satisfy:

$$ (D(n+2)-(D(n+3)+1)) (D(n+1)-(D(n+2)+1))(a_1 D(n+1) + a_2 D(n+2)+a_3 D(n+3)+a_4)\equiv 0 \pmod p \qquad (1)$$

for constants $a_i$. The constants $a_i$ depend on $p$ only and are determined by the first few values of $D(i)$ for which the other factors don't vanish. We don't know closed form for $a_i$.

The numerical evidence is 50 primes greater than $10^6$ and $10^3$ triples per each prime. Also several primes greater than $10^{50}$ were successfully tested.

Q1 Is the above identity true?

Remarks: Let $A=2^n \bmod p$. Then $a(n)=A^{p-1} \bmod p^2$ and $a(n+1)=(2A \bmod p)^{p-1} \bmod p^2$

Finding $n$ given $a(n)$ will break the discrete logarithm, which would be major result.

Answer 1

Using the notation from my answer to the previous question, we have $$D(n+1)−(D(n+2)+1)\not\equiv 0\pmod{p}$$ if and only if $g_0(n+2) = 2 g_0(n+1) - p$ and $g_1(n+2) \equiv 2 g_1(n+1) + 1\pmod{p}$, in which case $$D(n+2) \equiv -(n+2 + \frac{2 g_1(n+1) + 1}{2 g_0(n+1)c})\equiv D(n+1) - 1 - \frac{1}{2 g_0(n+1)c} \pmod{p}.$$ Similarly, $$D(n+2)−(D(n+3)+1)\not\equiv 0\pmod{p}$$ if and only if $$D(n+3) \equiv D(n+2) - 1 - \frac{1}{2 g_0(n+2)c}\equiv D(n+2) - 1 - \frac{1}{4 g_0(n+1)c}\pmod{p}.$$ If both these incongruences hold, then $$D(n+1) - 3D(n+2) + 2D(n+3) + 1 \equiv 0\pmod{p}$$ and so we can always take $(a_1,a_2,a_3,a_4)=(1,-3,2,1)$.