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%I #65 Feb 07 2020 23:56:42
%S 2,4,8,10,12,15,19,20,24,25
%N Smallest k such that k! can be represented as the sum of the n-th powers of two or more distinct primes; or -1 if no such k exists.
%C If a(10)..a(14) exist then a(10) > 26, a(11) > 28, a(12) > 30, a(13) > 32, a(14) > 33.
%C From _Jon E. Schoenfield_, Jun 07 2019: (Start)
%C If such a number k exists for n=8, then a(8) > 23.
%C 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:
%C - if 17 is one of the 95 primes, j must be either 39 or 56;
%C - otherwise, j must be either 31 or 48.
%C If such a number k exists for n=12, a(12) > 43.
%C (For proofs and additional notes, see the Links.) (End)
%H dxdy forum, <a href="https://dxdy.ru/post1395207.html">post</a> (in Russian).
%H Dmitry Kamenetsky, <a href="/A308357/a308357_1.txt">Solutions for n >= 8</a>
%H Carlos Rivera, <a href="https://www.primepuzzles.net/puzzles/puzz_964.htm">Puzzle 964: A308357</a>, The Prime Puzzles and Problems Connection.
%H Carlos Rivera, <a href="https://www.primepuzzles.net/puzzles/puzz_988.htm">Puzzle 988: Another puzzle about factorials</a>, The Prime Puzzles and Problems Connection.
%H Jon E. Schoenfield, <a href="/A308357/a308357.txt">A proof that a(8) > 23 (if a(8) != -1) and additional notes</a>
%e a(0) = 2, because 2! = 2 = 2^0 + 3^0.
%e a(1) = 4, because 4! = 24 = 11^1 + 13^1.
%e 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.
%e a(3) = 10, because 10! = 3628800 = 5^3 + 19^3 + 29^3 + 37^3 + 47^3 + 151^3.
%e 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.
%e 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.
%e 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}.
%e 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}.
%e Note that these are the smallest k for which such a representation is possible.
%Y Cf. A308459, A079612.
%K nonn,more,hard
%O 0,1
%A _Dmitry Kamenetsky_, May 21 2019
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