A333841 Integers n such that n! = x^2 + y^3 + z^4 where x, y and z are nonnegative integers, is soluble.

0, 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24

Offset: 1

Author

Altug Alkan, Aug 14 2020

Keywords

• nonn
• more

Examples

			6! = 11^2+7^3+4^4; 8! = 192^2+15^3+3^4;  9! = 443^2+55^3+4^4; 10! = 1888^2+40^3+4^4; 11! = 5896^2+172^3+16^4, so 6, 8, 9, 10 and 11 are in the sequence. - _R. J. Mathar_, Dec 15 2020
		

Crossrefs

Cf. A123053, A267414, A337046.

Formula

{k: k! in A123053}. - R. J. Mathar, Dec 15 2020