A003342 Numbers that are the sum of 8 positive 4th powers.

8, 23, 38, 53, 68, 83, 88, 98, 103, 113, 118, 128, 133, 148, 163, 168, 178, 183, 193, 198, 213, 228, 243, 248, 258, 263, 278, 293, 308, 323, 328, 338, 343, 353, 358, 368, 373, 388, 403, 408, 418, 423, 433, 438, 453, 468, 483, 488, 498, 503, 518, 533, 548, 563, 568

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Examples

			From _David A. Corneth_, Aug 04 2020: (Start)
5396 is in the sequence as 5396 = 1^4 + 1^4 + 4^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4.
8789 is in the sequence as 8789 = 5^4 + 5^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4 + 7^4.
12469 is in the sequence as 12469 = 1^4 + 3^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 10^4. (End)
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
• Eric Weisstein's World of Mathematics, Biquadratic Number.

Crossrefs

Cf. A000583, A003331, A003341, A003343, A003353, A345577, A345833.

Programs

Mathematica

Select[Range[500], AnyTrue[PowersRepresentations[#, 8, 4], First[#]>0&]&] (* Jean-François Alcover, Jul 18 2017 *)

Python

from itertools import combinations_with_replacement as mc
from sympy import integer_nthroot
def iroot4(n): return integer_nthroot(n, 4)[0]
def aupto(lim):
    pows4 = set(i**4 for i in range(1, iroot4(lim)+1) if i**4 <= lim)
    return sorted(t for t in set(sum(c) for c in mc(pows4, 8)) if t <= lim)
print(aupto(568)) # Michael S. Branicky, Aug 23 2021