A346365 Numbers that are the sum of six fifth powers in exactly ten ways.
55302546200, 89999127392, 96110537743, 104484239200, 120492759200, 121258798144, 127794946400, 133364991375, 135030535200, 136156575744, 151305014432, 155434423925, 174388570400, 177099008000, 179272687000, 182844944832, 184948721056, 187873845500
Offset: 1
Keywords
Examples
55302546200 = 34^5 + 38^5 + 50^5 + 57^5 + 95^5 + 136^5 = 23^5 + 49^5 + 61^5 + 69^5 + 107^5 + 131^5 = 24^5 + 37^5 + 63^5 + 81^5 + 104^5 + 131^5 = 21^5 + 35^5 + 60^5 + 94^5 + 100^5 + 130^5 = 57^5 + 60^5 + 71^5 + 75^5 + 109^5 + 128^5 = 19^5 + 37^5 + 56^5 + 96^5 + 104^5 + 128^5 = 35^5 + 41^5 + 53^5 + 69^5 + 115^5 + 127^5 = 16^5 + 49^5 + 53^5 + 83^5 + 112^5 + 127^5 = 35^5 + 37^5 + 40^5 + 88^5 + 119^5 + 121^5 = 11^5 + 24^5 + 71^5 + 104^5 + 109^5 + 121^5 so 55302546200 is a term.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..57
Programs
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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])
Comments