A346361 Numbers that are the sum of six fifth powers in exactly six ways.

287718651, 553545456, 746783675, 972232800, 1005620508, 1040741042, 1070652352, 1074892544, 1182426366, 1197332400, 1243267146, 1317183650, 1364866263, 1387455091, 1429663400, 1498160992, 1529189818, 1554833117, 1558594400, 1610298901, 1623782765, 1627228231

Offset: 1

Author

David Consiglio, Jr., Jul 13 2021

Keywords

• nonn

Comments

Differs from A345720 at term 10 because 1184966816 = 15^5 + 24^5 + 27^5 + 38^5 + 39^5 + 63^5 = 2^5 + 28^5 + 36^5 + 36^5 + 42^5 + 62^5 = 4^5 + 24^5 + 38^5 + 38^5 + 40^5 + 62^5 = 21^5 + 32^5 + 37^5 + 41^5 + 45^5 + 60^5 = 8^5 + 14^5 + 34^5 + 40^5 + 52^5 + 58^5 = 11^5 + 17^5 + 22^5 + 49^5 + 51^5 + 56^5 = 11^5 + 16^5 + 22^5 + 52^5 + 52^5 + 53^5.

Examples

			287718651 is a term because 287718651 = 10^5 + 11^5 + 20^5 + 22^5 + 30^5 + 48^5 = 8^5 + 10^5 + 21^5 + 27^5 + 27^5 + 48^5 = 3^5 + 6^5 + 25^5 + 30^5 + 30^5 + 47^5 = 9^5 + 10^5 + 13^5 + 26^5 + 37^5 + 46^5 = 6^5 + 9^5 + 14^5 + 31^5 + 35^5 + 46^5 = 10^5 + 11^5 + 12^5 + 23^5 + 41^5 + 44^5.
		

Links

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

Crossrefs

Cf. A345720, A345818, A346283, A346360, A346362.

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 == 6])
    for x in range(len(rets)):
        print(rets[x])