A345833 Numbers that are the sum of eight fourth powers in exactly one ways.

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, 328, 338, 353, 368, 403, 408, 418, 433, 468, 483, 488, 498, 568, 578, 593, 608, 632, 643, 647, 648, 658, 662, 663, 673, 677, 692, 707, 708, 712

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A003342 at term 26 because 263 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4.

Examples

			23 is a term because 23 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003342, A345783, A345823, A345834, A345843, A346326.

Programs

Python

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1, 1000)]
for pos in cwr(power_terms, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 1])
    for x in range(len(rets)):
        print(rets[x])