A344196 Numbers that are the sum of six fifth powers in ten or more ways.

55302546200, 89999127392, 96110537743, 104484239200, 120492759200, 121258798144, 127794946400, 133364991375, 135030535200, 136156575744, 151305014432, 155434423925, 174388570400, 177099008000, 179272687000, 180336745600, 182844944832, 184948721056, 187873845500

Offset

1,1

Links

Sean A. Irvine, Table of n, a(n) for n = 1..61

Example

89999127392 is a term because 89999127392 = 4^5 + 36^5 + 39^5 + 40^5 + 90^5 + 153^5 = 8^5 + 21^5 + 90^5 + 109^5 + 119^5 + 135^5 = 8^5 + 28^5 + 98^5 + 102^5 + 104^5 + 142^5 = 10^5 + 38^5 + 74^5 + 102^5 + 118^5 + 140^5 = 13^5 + 51^5 + 64^5 + 98^5 + 112^5 + 144^5 = 18^5 + 44^5 + 66^5 + 98^5 + 112^5 + 144^5 = 18^5 + 52^5 + 72^5 + 78^5 + 118^5 + 144^5 = 28^5 + 60^5 + 63^5 + 65^5 + 124^5 + 142^5 = 36^5 + 53^5 + 62^5 + 63^5 + 129^5 + 139^5 = 39^5 + 41^5 + 64^5 + 91^5 + 98^5 + 149^5.

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 >= 10])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345567, A345643, A345723, A346365.

Sequence in context: A034654 A233553 A273928 * A346365 A295042 A320880

Adjacent sequences: A344193 A344194 A344195 * A344197 A344198 A344199

Keyword

nonn

Author

David Consiglio, Jr., Jun 25 2021

Status

approved