Largest Prime Removing Digits From Left-Hand End

Recently on Fermat's Library in Twitter, the following curiosity was posted:


What if we tried the same but from the left-hand end? For example, "113" would be a candidate because:

113 is prime
13 is prime
3 is prime

What's the largest of such a value? Is the a max one? Well let's leave the digit "zero" aside for now since we're looking to have unique numbers at each step of the sequence.
There is actually a Max value for that! Here it is:

357686312646216567629137

Code to find it is down below - it uses BigInteger and Miller-Rabin for quick primality test. Also, DFS with pruning. Equally interesting as the twitter post. Cheers, ACC.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Numerics;
namespace FermatLibraryProblem
{
 class Program
 {
 static void Main(string[] args)
 {
 BigInteger max = 0;
 Process(0, 1, ref max);
 Console.WriteLine();
 Console.WriteLine("Solution: {0}", max);
 }
 public static void Process(BigInteger number, BigInteger multiplier, ref BigInteger max)
 {
 if (number != 0 && !IsPrimeMillerRabin(number)) return;
 if (number > max)
 {
 max = number;
 Console.WriteLine("New Max: {0}", max);
 }
 for (int i = 1; i < 10; i++)
 {
 Process(multiplier * i + number, 10 * multiplier, ref max);
 }
 }
 public static bool IsPrimeMillerRabin(BigInteger n)
 {
 //It does not work well for smaller numbers, hence this check
 int SMALL_NUMBER = 1000;
 if (n <= SMALL_NUMBER)
 {
 return IsPrime(n);
 }
 int MAX_WITNESS = 500;
 for (long i = 2; i <= MAX_WITNESS; i++)
 {
 if (IsPrime(i) && Witness(i, n) == 1)
 {
 return false;
 }
 }
 return true;
 }
 public static BigInteger SqRtN(BigInteger N)
 {
 /*++
 * Using Newton Raphson method we calculate the
 * square root (N/g + g)/2
 */
 BigInteger rootN = N;
 int count = 0;
 int bitLength = 1;
 while (rootN / 2 != 0)
 {
 rootN /= 2;
 bitLength++;
 }
 bitLength = (bitLength + 1) / 2;
 rootN = N >> bitLength;
 BigInteger lastRoot = BigInteger.Zero;
 do
 {
 if (lastRoot > rootN)
 {
 if (count++ > 1000) // Work around for the bug where it gets into an infinite loop
 {
 return rootN;
 }
 }
 lastRoot = rootN;
 rootN = (BigInteger.Divide(N, rootN) + rootN) >> 1;
 }
 while (!((rootN ^ lastRoot).ToString() == "0"));
 return rootN;
 }
 public static bool IsPrime(BigInteger n)
 {
 if (n <= 1)
 {
 return false;
 }
 if (n == 2)
 {
 return true;
 }
 if (n % 2 == 0)
 {
 return false;
 }
 for (int i = 3; i <= SqRtN(n) + 1; i += 2)
 {
 if (n % i == 0)
 {
 return false;
 }
 }
 return true;
 }
 private static int Witness(long a, BigInteger n)
 {
 BigInteger t, u;
 BigInteger prev, curr = 0;
 BigInteger i;
 BigInteger lln = n;
 u = n / 2;
 t = 1;
 while (u % 2 == 0)
 {
 u /= 2;
 t++;
 }
 prev = BigInteger.ModPow(a, u, n);
 for (i = 1; i <= t; i++)
 {
 curr = BigInteger.ModPow(prev, 2, lln);
 if ((curr == 1) && (prev != 1) && (prev != lln - 1)) return 1;
 prev = curr;
 }
 if (curr != 1) return 1;
 return 0;
 }
 }
}

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