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

1,1

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.

Links

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

Example

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.

Prog

(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])

Crossrefs

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

Sequence in context: A344518 A344519 A345720 * A183737 A234807 A204055

Adjacent sequences: A346358 A346359 A346360 * A346362 A346363 A346364

Keyword

nonn

Author

David Consiglio, Jr., Jul 13 2021

Status

approved