BigInteger.LucasSequenceHelper C# (CSharp) Method

        private static BigInteger[] LucasSequenceHelper(BigInteger P, BigInteger Q,
                                                        BigInteger k, BigInteger n,
                                                        BigInteger constant, int s)
        {
                BigInteger[] result = new BigInteger[3];

                if((k.data[0] & 0x00000001) == 0)
                        throw (new ArgumentException("Argument k must be odd."));

                int numbits = k.bitCount();
                uint mask = (uint)0x1 << ((numbits & 0x1F) - 1);

                // v = v0, v1 = v1, u1 = u1, Q_k = Q^0

                BigInteger v = 2 % n, Q_k = 1 % n,
                           v1 = P % n, u1 = Q_k;
                bool flag = true;

                for(int i = k.dataLength - 1; i >= 0 ; i--)     // iterate on the binary expansion of k
                {
                        //Console.WriteLine("round");
                        while(mask != 0)
                        {
                                if(i == 0 && mask == 0x00000001)        // last bit
                                        break;

                                if((k.data[i] & mask) != 0)             // bit is set
                                {
                                        // index doubling with addition

                                        u1 = (u1 * v1) % n;

                                        v = ((v * v1) - (P * Q_k)) % n;
                                        v1 = n.BarrettReduction(v1 * v1, n, constant);
                                        v1 = (v1 - ((Q_k * Q) << 1)) % n;

                                        if(flag)
                                                flag = false;
                                        else
                                                Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);

                                        Q_k = (Q_k * Q) % n;
                                }
                                else
                                {
                                        // index doubling
                                        u1 = ((u1 * v) - Q_k) % n;

                                        v1 = ((v * v1) - (P * Q_k)) % n;
                                        v = n.BarrettReduction(v * v, n, constant);
                                        v = (v - (Q_k << 1)) % n;

                                        if(flag)
                                        {
                                                Q_k = Q % n;
                                                flag = false;
                                        }
                                        else
                                                Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);
                               }

                               mask >>= 1;
                        }
                        mask = 0x80000000;
                }

                // at this point u1 = u(n+1) and v = v(n)
                // since the last bit always 1, we need to transform u1 to u(2n+1) and v to v(2n+1)

                u1 = ((u1 * v) - Q_k) % n;
                v = ((v * v1) - (P * Q_k)) % n;
                if(flag)
                        flag = false;
                else
                        Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);

                Q_k = (Q_k * Q) % n;


                for(int i = 0; i < s; i++)
                {
                        // index doubling
                        u1 = (u1 * v) % n;
                        v = ((v * v) - (Q_k << 1)) % n;

                        if(flag)
                        {
                                Q_k = Q % n;
                                flag = false;
                        }
                        else
                                Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);
                }

                result[0] = u1;
                result[1] = v;
                result[2] = Q_k;

                return result;
        }