Find the Missing Members of this Powered Sequence

There are 4 missing members in this small series...labeled ???.

Hint 1;

When you need assistance, seek help from higher authority.

Hint 2:

Higher authority can grant you higher powers!

Complete the series.

$$191$$, $$426$$, $$931$$, $$???$$, $$???$$, $$646$$, $$971$$, $$???$$, $$???$$

• If nobody gets it by tomorrow, I will drop a Hint. – Uvc Jun 9 at 17:01
• Going to provide the first hint now. – Uvc Jun 10 at 10:01
• Second hint is being given. – Uvc Jun 11 at 2:00

The sequence is:

191, 426, 931, 666, 555, 646, 971, 486, 111

Explanation:

For the nth number:

The first digit is: $$n^2 \bmod 10$$.
The second digit is: $$10 - (n^3 \bmod 10)$$.
The third digit is: $$n^4 \bmod 10$$.

Example: eighth number

First digit is: $$8^2 \bmod 10 = 64 \bmod 10 = 4$$
Second digit is: $$10 - (8^3 \bmod 10) = 10 - (512 \bmod 10) = 10 - 2 = 8$$
Third digit is: $$8^4 \bmod 10 = 4096 \bmod 10 = 6$$
Final number: $$486$$

• Got it......... – Uvc Jun 11 at 2:35

191, 426, 931, 626, 591, 646, 971, 486, 191

because

First digits are the square numbers modulo 10 (1, 4, 9, 6, 5, 6, 9, 4, 1) and last digits alternate between 1 and 6. Respectively, the sum of all the digits add up to 11, 12, 13, 14, 15, 16, 17, 18 and cycle back to 11 since the last term is the same as the starting one.

• nice progress and approach, +1! – Omega Krypton Jun 9 at 9:09
• any ideas? mine wont work... – Omega Krypton Jun 9 at 9:23
• All I can say is that you are on the right track partially. – Uvc Jun 9 at 9:28

191, 426, 931, ???, ???, 646, 971, ???, ???

After looking for few minute I found pattern as

1.Sum of digit in series as 11,12,13, ???, ???,16,17,???, ???

2.the first digit: square of series 1,4,9,1(6),2(5),3(6),4(9),6(4),8(1)

3.the last digit is flipping between 1& 6

So using 1&2&3 pattern :626,591,486,991

• Not right..will post a hint tomorrow if nobody gets it by that time – Uvc Jun 9 at 18:14