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

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

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 $D(n) \mod p=D(n+1)+1 \mod p$.

>Q1 Are there other functional relations between $g,n,a(n),D(n)$?

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

>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