A346359 Numbers that are the sum of six fifth powers in exactly four ways.

12047994, 20646208, 21017489, 21300963, 21741819, 24993485, 27669050, 28576064, 30193856, 30785920, 35480456, 35735194, 36082750, 37303264, 39035975, 46814942, 47963291, 50047062, 50724345, 52987561, 53076800, 53606848, 55101101, 56766906, 57969327, 58125980

Offset: 1

Author

David Consiglio, Jr., Jul 13 2021

Keywords

• nonn

Comments

Differs from A345718 at term 23 because 54827300 = 4^5 + 7^5 + 21^5 + 22^5 + 23^5 + 33^5 = 5^5 + 10^5 + 15^5 + 20^5 + 28^5 + 32^5 = 1^5 + 14^5 + 16^5 + 19^5 + 28^5 + 32^5 = 4^5 + 11^5 + 13^5 + 22^5 + 29^5 + 31^5 = 5^5 + 6^5 + 19^5 + 20^5 + 29^5 + 31^5.

Examples

			12047994 is a term because 12047994 = 7^5 + 9^5 + 12^5 + 14^5 + 17^5 + 25^5 = 5^5 + 10^5 + 13^5 + 15^5 + 16^5 + 25^5 = 1^5 + 1^5 + 3^5 + 4^5 + 21^5 + 24^5 = 4^5 + 6^5 + 15^5 + 15^5 + 21^5 + 23^5.
		

Links

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

Crossrefs

Cf. A344519, A345718, A345816, A346281, A346358, A346360.

Programs

Python

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**5 for x in range(1, 1000)]
for pos in cwr(power_terms, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 4])
    for x in range(len(rets)):
        print(rets[x])