Crossnumber Puzzle

Question

Across:

1. Sum of consecutive integer powers of 21 Across
7. Prime number, not all of whose digits are prime
8. Number that is coprime with 13 Down
9. Sum of three consecutive primes
10. Noble gas atomic number
12. 5 more than 9 Across
14. US coin denomination
15. Permutable prime
17. Fibonacci number
20. Power of 2 minus 1
21. Last two digits of 16 Down

Down:

1. 2 Down in base 21 Across
2. Leap year
3. Number not divisible by any of its digits
4. Number whose cube consists of three digits, each occurring twice
5. Palindromic number
6. 1 Across mod 7 Across
10. All 1-digit factors of 20 Across (excluding 1), in ascending order
11. 1 greater than 4 Down
13. Number that is less than 8 Across
16. CCCXV
18. Sum of digits of 8 Across
19. Can be expressed as sum of two squares, neither equal to 1, in two ways

Other notes:

• Heavy lines indicate the beginnings and endings of entries.
• No answers have leading zeros.

Answer 1

Solution:

The first steps are clear.

16D: 315

21A: 15

17A: 317811

20A: 4095 (by 10D, this number is a multiple of 7)

10D: 3579

14A: 25

10A: 36

4D: 62 (by 11D, this number is between 59 and 68)

11D: 63

15A: 373

5D: 117711

13D: 134 (since it's smaller than 8A)

8A: 217 (by 18A, the sum of digits of this number is 10)

9A: 71 (by 12A, this number is between 65 and 74)

12A: 76

Here things become more tricky.

By 1D, an argument mod 25 tells us that the starting digit of 7A is 2 or 7.
Assume 7A starts with 2:

2D: 1612

1D: 727

6D: 176

At this point, I bruteforced the remaining two digits and saw that there was no solution.

Thus we conclude that

7A starts with 7, and hence

7A: 797

Now there are again two possibilities:

1D: 477 or 877.
Assume 1D: 477, then:

2D: 1012

7A: 709 (by 3D, the last digit cannot be 1)

Bruteforcing the remaining two digits again gives no solution.

Thus we get

1D: 877

2D: 1912

7A: 797

And finally

bruteforcing the remaining two digits gives the above answer (note that 6D cannot start with 0).

What I cannot verify:

1A: Sum of consecutive integer powers of 21 Across
I don't understand what this means...

EDIT:

I see, the sentence in 1A above is refering to the fact that $813616 = 15^0 + 15^1 + \dots + 15^5$. Thus my procedure above could be simplified a lot.

Answer 2

Partial answer:

[4 down] 62 (= 238328^1/3)

[16 down] 315

[19 down] 85

[20 across] 4095 (= 2^12 - 1)

[21 across] 15

[17 across] 317811 (= 6-digit Fibonacci number ???8?1)