A345631 Numbers that are the sum of seven fifth powers in nine or more ways.
110276376, 124732805, 127808693, 130298618, 134581976, 188116743, 189642309, 202274051, 202686274, 203343582, 219063107, 230909843, 233137574, 233549568, 234250752, 235438301, 244250335, 251138524, 252277376, 253480833, 254017026, 254380543
Offset: 1
Keywords
Examples
124732805 is a term because 124732805 = 3^5 + 18^5 + 22^5 + 22^5 + 24^5 + 27^5 + 39^5 = 4^5 + 15^5 + 17^5 + 21^5 + 29^5 + 34^5 + 35^5 = 5^5 + 14^5 + 17^5 + 24^5 + 25^5 + 35^5 + 35^5 = 7^5 + 8^5 + 17^5 + 26^5 + 29^5 + 34^5 + 34^5 = 7^5 + 10^5 + 12^5 + 31^5 + 31^5 + 32^5 + 32^5 = 7^5 + 12^5 + 23^5 + 24^5 + 24^5 + 26^5 + 39^5 = 7^5 + 14^5 + 22^5 + 22^5 + 23^5 + 28^5 + 39^5 = 16^5 + 25^5 + 25^5 + 28^5 + 28^5 + 28^5 + 35^5 = 20^5 + 24^5 + 24^5 + 25^5 + 25^5 + 32^5 + 35^5.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..10000
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, 7): tot = sum(pos) keep[tot] += 1 rets = sorted([k for k, v in keep.items() if v >= 9]) for x in range(len(rets)): print(rets[x])