A344730 Numbers that are the sum of three fourth powers in exactly seven ways.

779888018, 12478208288, 33038379458, 63170929458, 114872872562, 199651332608, 329296962722, 393006728738, 419200136082, 487430011250, 528614071328, 959702600738, 1010734871328, 1369390032738, 1502549262242, 1525400097858, 1653983981762, 1668273965442, 1756039197458, 1793250582818, 1837965960992, 1912768493202

Offset: 1

Author

David Consiglio, Jr., May 27 2021

Keywords

• nonn

Comments

Differs from A344729 at term 2 because 5745705602 3^4+ 230^4+ 233^4 = 25^4+ 218^4+ 243^4 = 43^4+ 207^4+ 250^4 = 58^4+ 197^4+ 255^4 = 85^4+ 177^4+ 262^4 = 90^4+ 173^4+ 263^4 = 102^4+ 163^4+ 265^4 = 122^4+ 145^4+ 267^4

Examples

			779888018 is a term because 779888018 = 3^4+ 139^4+ 142^4 = 9^4+ 38^4+ 167^4 = 14^4+ 133^4+ 147^4 = 43^4+ 114^4+ 157^4 = 47^4+ 111^4+ 158^4 = 63^4+ 98^4+ 161^4 = 73^4+ 89^4+ 162^4
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..86

Crossrefs

Cf. A344648, A344729, A344738, A344923, A345085.

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, 3):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 7])
for x in range(len(rets)):
    print(rets[x])