
COMMENTS

If a(10)..a(14) exist then a(10) > 26, a(11) > 28, a(12) > 30, a(13) > 32, a(14) > 33.
From Jon E. Schoenfield, Jun 07 2019: (Start)
If such a number k exists for n=8, then a(8) > 23.
If a set of distinct primes whose 8th powers sum to 24! exists (i.e., if a(8)=24), it must consist of exactly 96 primes: p=5 and exactly 95 primes p > 5. Additionally, if we let j be the number of primes p satisfying p^8 mod 17 = 1 among those 95 primes > 5, it can be shown that:
 if 17 is one of the 95 primes, j must be either 39 or 56;
 otherwise, j must be either 31 or 48.
If such a number k exists for n=12, a(12) > 43.
(For proofs and additional notes, see the Links.) (End)


EXAMPLE

a(0) = 2, because 2! = 2 = 2^0 + 3^0.
a(1) = 4, because 4! = 24 = 11^1 + 13^1.
a(2) = 8, because 8! = 40320 = 2^2 + 3^2 + 5^2 + 7^2 + 11^2 + 13^2 + 17^2 + 19^2 + 23^2 + 29^2 + 41^2 + 59^2 + 181^2.
a(3) = 10, because 10! = 3628800 = 5^3 + 19^3 + 29^3 + 37^3 + 47^3 + 151^3.
a(4) = 12, because 12! = 479001600 = 3^4 + 5^4 + 7^4 + 11^4 + 17^4 + 19^4 + 29^4 + 31^4 + 37^4 + 47^4 + 53^4 + 59^4 + 73^4 + 79^4 + 97^4 + 131^4.
a(5) = 15, because 15! = 13^5 + 17^5 + 19^5 + 31^5 + 37^5 + 41^5 + 53^5 + 61^5 + 89^5 + 97^5 + 139^5 + 163^5 + 199^5 + 241^5.
a(6) = 19, because 19! is the sum of the 6th powers of the primes in {3, 7, 17, 23, 37, 43, 47, 53, 61, 71, 73, 79, 89, 101, 103, 107, 113, 127, 137, 157, 167, 193, 211, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463}.
a(7) = 20, because 20! is the sum of the 7th powers of the primes in {5, 13, 31, 43, 59, 67, 71, 83, 97, 103, 109, 113, 137, 149, 167, 179, 181, 191, 193, 197, 227, 229, 233, 239, 241, 257, 263, 269, 277, 281, 283, 293, 311, 313, 317, 331}.
Note that these are the smallest k for which such a representation is possible.
