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

Author

David Consiglio, Jr., Jun 25 2021

Keywords

• nonn

Examples

			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
so 89999127392 is a term.
		

Links

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

Crossrefs

Cf. A345567, A345643, A345723, A346365.

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