Let $p$ be prime and $g,n$ integers.

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

Some properties of $a(n)$:

1. $a(n)$ is periodic with period divisor of $p-1$.
2. The multiplicative order of $a(n)$ modulo $p^2$ is $p$.
3. Let $D(n)$ be the discrete logarithm of $a(n)$, i.e. given $p$, $g$, $a(n)$ we have $g^{D(n)} \mod p^2=a(n)$. We can efficiently compute $D(n)=k(p-1)$ via p-adic logarithms.
4. Let $g=2$. Experimentally with high probability we have $D(n) \bmod p=D(n+1)+1 \bmod p$.

Q1 Are there other functional relations between $g$, $n$, $a(n)$, $D(n)$? We believe that finding $n$ given $p,g,a(n)$ will solve the discrete logarithm modulo $p$, which is a major result.

Q2 For $g=2$, when do we have $D(n) \bmod p=D(n+1)+1 \bmod p$?

Q3 What is the intuition for efficiently computing $D(n)$ for period divisor of $p-1$?

sagemath code follows, one can run it in a browser:

def seqanp2(p,g,n):
    """
    a(n)=(g^n mod p)^(p-1) mod p^2
    """
    try:  g=lift(g)
    except:  pass
    r1=lift((Integers(p)(g))**n)
    K2=Integers(p**2)
    res=K2(r1)**((p-1))
    return res

def solveseqan(p,g,a):
    """
    g^res =a(n)  mod p^2
    """
    try:  g=lift(g)
    except:  pass
    try:  a=lift(a)
    except:  pass
    K=Qp(p,2)
    t=lift(K(a).log()/K(g).log() )
    res=(p-1)*(p-t%p)
    return res

set_random_seed(1)

p=next_prime(10**20);
K2=Integers(p**2);
g=K2(2) 
n0=randint(2,p-2)
r1=seqanp2(p,g,n0);r2=seqanp2(p,g,n0+1);
s1=solveseqan(p,g,r1);s2=solveseqan(p,g,r2)

print(g**s1==r1,g**s2==r2,seqanp2(p,g,n0)==seqanp2(p,g,n0+p-1)) #True True True