Finding n mod p^(D-1) given A=g^n mod p^D

[Georgi Guninski <gguninski@gmail.com>]

We asked this on mathoverlow [1], but didn't get answer.

Let p be prime and n,g,D integers. Let A=g^n mod p^D

Let dlog(p,g,A,D)=log(A+O(p^D))/log(g+O(p^D)).

Conjecture 1: dlog(p,g,A,D) mod p^(D-1) = n mod p^(D-1)

dlog is efficiently computable.

We verified it with hundreds of large tuples (p,g,D,n).

It is also related to the discrete logarithm of A in base g
modulo p^D.

Is the conjecture true?

[1]: https://mathoverflow.net/questions/391223/p-adic-logarithms-with-fixed-precision

Code:

/*
discrete logarithm modulo p^(D-1)
A=g^n mod p^D

dlog(p,g,A,D) mod p^(D-1) = n mod p^(D-1)

Author Georgi Guninski
*/

{
dlog1(p,g,a,D=2)=lift(log(lift(a)+O(p^D))/log(lift(g)+O(p^D)));
}

{
tt()=
D=3;
setrand(1);
p=nextprime(10^18);X0=random(p^D);g=Mod(2,p^D);a=g^X0;
X1=dlog1(p,g,a,D);
\\print(X1,X0);
print([(X1-X0)%p^(D-1)]);
}
tt()